419 + Interview Questions in MODERN-PHYSICS IN GENERAL AWARENESS Page 8 InterviewSolution

351.	The electron in a hydrogen atom make a transtion `n_(1) rarr n_(2)` where `n_(1) and n_(2)` are the priocipal quantum number of the two states . Assume the Bohr model to be valid . The time period of the electron in the initial state is eight time that in the state . THe possible values of `n_(1) and n_(2)` areA. `n_(1) = 4 , n_(2) = 2 `B. n_(1) = 8 , n_(2) = 2`C. n_(1) = 8 , n_(2) = 1 `D. n_(1) = 6 , n_(2) = 3`
Answer» Correct Answer - A::D The time period of the electron in a bohr is given by `T = (2 pi r )/(nu)` Since the for the bohr orbit , `m nu r = n (h//2 pi) ` , the time period because `T = (2 pi r)/(nh//(2 tau m r )) = ((4 pi ^(2) m)/(nh))` since the radius of the orbit r depends on n, we replace r bohr radius of a hydrogen atom is `r = n^(2) ((h^(2)epsilon _(0))/(tau me^(2))` Hence , T = ((4 pi ^(2) m)/(nh)) (( n^(2) h^(4) epsilon_(0)^(2))/(pi ^(2) m^(2) e^(4))) = n^(2) (( 4h^(3) epsilon_(0)^(2))/(me^(4)))` `for two orbit `T_(1))/(T_(2)) = (( n_(1))/(n_(2))) ^(2) ` it is given that `T_(1) //T_(2) = 8 ,` heanse `n_(1)//n_(2) = 2 `

Discussion

352.	Hydrogen atom is exited from ground state to another state with principal quantum number equal to `4` Then the number of spectral linear in the emission spectra will beA. `2`B. `3`C. `5`D. `6`
Answer» Correct Answer - D The possible number of the spectral lines is given `= (n (n - 1))/(2) = (4 (4 - 1))/(2) = 6`

Discussion

353.	de-Broglie wavelength of an electron in the nth Bohr orbit is `lambda_(n)` and the angular momentum is `J_(n)` thenA. `J_(n)proplambda_(n)`B. `lambda_(n)prop(1)/(J_(n))`C. `lambda_(n)propJ_(n)^(2)`D. None of these
Answer» Correct Answer - A

Discussion

354.	Find the maximum angular speed of the electron of a hydrogen atoms in a statoonary orbitA. `6.2xx10^(5)rad//s`B. `4.1xx10^(16) rad//s`C. `2.4xx10^(10) rad//s`D. `9.2xx10^(10) rad//s`
Answer» Correct Answer - B

Discussion

355.	A star initially has `10^40` deuterons. It produces energy via the processes `_1H^2+_1H^2rarr_1H^3+p` and `_1H^2+_1H^3rarr_2He^4+n`. If the average power radiated by the star is `10^16` W, the deuteron supply of the star is exhausted in a time of the order of (a) `10^6s` (b) `10^8s` (c) `10^12s` The masses of the nuclei are as follows `M(H^2)=2.014` amu, `M(n)=1.008` amu, `M(p)=1.007` amu,`M(He^4)=4.001`amu
Answer» Correct Answer - C The given reactions are `_1H^2+_1H^2rarr_1H^3+p` `_1H^2+_1H^3rarr_2He^4+n` implies `3_1H^2rarr_2He^4+n+p` Mass defect, `Deltam=(3xx2.014-4.001-1.007-1.008)` amu `=0.026` amu Energy released `=0.026 xx 931MeV` `=0.026xx931xx1.6xx10^-13J` `=3.87xx10^-12J` This is the energy produced by the comsumption of three deuteron atoms. :. Total energy released by `10^40` deuterons `=10^40/3xx3.87xx10^-12J=1.29xx10^28J` The average power radiated is `P=10^16W` or `10^16J//s`. Therefore, total time to exhaust all deuterons of the star will be `t=(1.29xx10^28)/(10^16)=1.29xx10^12s~~10^12s` :. The correct option is (c).

Discussion

356.	When photon of wavelength `lambda_(1)` are incident on an isolated shere supended by an insulated , the corresponding stopping potential is found to be `V`. When photon of wavelength `lambda_(2)` are used , the orresponding stopping potential was thrice the above value. If light of wavelength `lambda_(3)` is used , carculate the stopping potential for this case.A. `(hc)/(e)((1)/(lambda_(3))+(1)/(2lambda_(2))-(1)/(lambda_(1)))`B. `(hc)/(e)((1)/(lambda_(3))+(1)/(2lambda_(2))-(1)/(2lambda_(1)))`C. `(hc)/(e)((1)/(lambda_(3))+(1)/(lambda_(2))+(1)/(lambda_(1)))`D. `(hc)/(e)((1)/(lambda_(3))-(1)/(lambda_(2))-(1)/(lambda_(1)))`
Answer» Correct Answer - B

Discussion

357.	A silver balll is suspended by a string in a vacuum chamber and ultraviolet light of wavelength `2000 Å` is directed at it. What electrical potential will the ball acquire as a result? Work function of silver is 4.3 eV.
Answer» Correct Answer - A `K_(max) (in eV) = E -W` `= (12375)/(2000) - 4.3` =1.9 eV Therefore, stopping potentia is 1.9 V.

Discussion

358.	The short-wavelength limit shifts by 26 pm when the operating voltage in an X-ray tube is increased to 1.5 times the original value. What was the original value of the operating voltage?
Answer» Correct Answer - A `26 pm = 0.26 Å` Now, 0.26 = (12475)/(V) - (12375)/(1.5 V).

Discussion

359.	x-rays are produced in an X-ray tube by electrons asselerated through an electric potential difference fo 50.0 kV. An electron makes three collisions in the target coming to rest and loses half its remaining kinetic energy in each of the first two collisions. Determine the wavelenght of the resulting photons. (Neglecting the recoil of the heavey target atoms).
Answer» Correct Answer - D `DeltaE_1 = 50% of 50 keV = 23keV` `:. Lambda_1 = (12375)/(25xx10^3) = 0.495 Å` = 49.5 pm `DeltaE_2 = 50% of 25 keV` =12.5 keV `:. Lambda_2 = (12375)/(12.5xx10^3) = 0.99 Å` = 99 pm.

Discussion

360.	if `lambda_(Cu)` is the wavelength of `K_alpha`, X-ray line fo copper (atomic number 29) and `lambda_(Mo)` is the wavelength of the `K_alpha` X-ray line of molybdenum (atomic number 42), then the ratio `lambda_(Cu)/lambda_(Mo)`is close to (a) 1.99 (b) 2.14 (c ) 0.50 (d) 0.48
Answer» Correct Answer - B `K_alpha transition takes place from n_1 = 2 to n_2 =1` `:. (1)/(lambda) = R(Z - b)^2 [(1)/((1))^2-(1)/((2))^2)]` For K-series, b=1 `:. (1)/(lambda) prop (Z-1)^2` `rArr (lambda_(Cu))/(lambda_Mo)) = ((Z_(mo) -1)^2/((Z_(Cu) - 1)^2 = (42-1)^2/(29-1)^2` `=(41xx41)/(28xx28) = (1681)/(784) = 2.144`

Discussion

361.	Wavelength of `K_alpha` line of an element is `lambda_0`. Find wavelength of `K_beta` - line for the same elemetn.
Answer» Correct Answer - A::B::C::D `(1)/(lambda) prop((1)/(n_1^2) - (1)/(n_2^2))` `:. (lambda_(K_beta)/(lambda_(K_alpha) = ((1//n_1^2 - 1 //n_2^2)/((1//n_1^2 -1//n_2^2)(K_Beta))` `=((1-1//4)/(1-1//9)) = (27)/(32)` `:. lambda_(K_beta) = (27)/(32) lambda_(K_alpha) = (27)/(32) lambda_0` `:.lambda_(K_beta) = (27)/(32) lambda_(K_alpha) = (27)/(32)lambda_0`

Discussion

362.	From what meterial is the anod of an X-ray tube made if the `K_alpha` line wavelength of thej characteristic spectrum is `0.76 Å`?
Answer» Correct Answer - A::D `(1)/(lambda_(K_alpha) = R(Z-1)^2 ((1)/(n_1^2) - (1)/(n_2^2))` or `(1)/(0.76xx10^(-10) = 1.097xx10^7 (Z-1)^2 (1 - (1)/(4)` Solving this equation, Z =41.

Discussion

363.	For a radioactive meterial , its activity `A` and rate of charge of its activity `R` are defined as `A = - (dN)/(dt) and R = (dA)/(dt) ` where `N(t)` is the number of nuclei at time I .Two radioactive source `P (mean life tau ) and Q(mean life 2 tau )` have the same activity at `t = 2 tau R_(p) and R_(Q)` respectively , if `(R_(p))/(R_(Q)) = (n)/(e)`
Answer» Correct Answer - B `R = (dA)/(dt) = -(d)/(dt) [(dN)/(dt)] = (d^(2) N)/(dt^(2) ) = (d^(2) (N_(0)e^(-lambda t)))/(dt^(2))` `:. R = N_(0) lambda^(2)e^(-lambda t) = (N_(0) lambda) lambda e^(lambda t) = A_(0)lambda e^(- lambda t)` [:. A_(0) = N_(0) lambda]` `:. (R_(p))/(R_(Q)) = (lambda_(p) e^(lambda_(0)t)/(lambda_(Q)e^(- lambda_(0)t) =) lambdap)/(lambda Q) xx e^(lambda_(0) t)/(e^(lambda_(0) t) = ( 2 pi e^((2 pi )/(2 pi))/((2 pi )/(e^(s))) = (2)/(e)` :. n = 2

Discussion

364.	The threshold wavelength for photoelectric from a material is `5200 Å` photoelectric will be emittillumnated with when this meterial is illiuminated with monecharnatic radiation from aA. `50 `watt infrared lampB. `1` - watt infra- red lampC. `50 `watt ubraviolet lampD. `1` - watt ahraviolet lamp
Answer» Correct Answer - C::D The thereshould wavelength also depands is `5200Å` for ejection of electrons the wavwlength of the liught should be loss thasn `5200Å` as that frequency increases and bance the energy of ncident photo increases ``U V light has less wavelength than `5200Å`

Discussion

365.	A nuclear power supplying electrical power to a villages uses a radioactive meterial of half life `T` year as the fiel . The amount of fuel at the beginning is such that the total power requirement of the village is `12.5%` of the electrical power available from the plate at that time if the plate is able to meet the total power needs of the village for a maximum period of `n T` year , then the value of `n ` is
Answer» Correct Answer - C `N_(0) overset(T) rarr (N_(0))/(2) overset(T) rarr (N_(0))/(4) overset(T) rarr (N_(0))/(8)` `100% 50% 25% 12.5%` therefore half life are required Therefore `n = 3`

Discussion

366.	In an agriculture experiment, a solution containing 1 mole of a radioactive meterial `(t_(1//2)=14.3 days)`was injected into the roots of a plants.the plant was allowed 70 hours to settle down and then activity eas measured in its fruit. If the activity measured was `1 mu Ci` what per cent of activity is transmitted from the root to the fruit in steady state?
Answer» Correct Answer - A::B `R_0=lambdaN=(0.693)/(14.3xx3600xx24)xx6.02xx10^23` per sec `=3.37xx10^17` per sec After 70 hours activity, `R=R_0e^(-lambdat)=(3.37xx10^17)e^(-(0.693//14.3xx24)(70))` `=2.92xx10^17` per sec In fruits activity was observed `1muCi` or `3.7xx10^4` per sec. Therefore, percentage of activity transmitted from root to the fruit. `=(3.7xx10^4)/(2.92xx10^17)xx100` `1.26xx10^-11%`

Discussion

367.	Light of wavelength `lambda_(ph)`falls on a plate a vacum teke as shown in the figure .The work function of the conducting meterial kept at a distance d from the cathon A petential different V is maximum between the electrodes if the minimum de Brogle waveleeength of the electrons passing through the anode is `lambda_(e) ` which of the following statement (s) is (are) true? A. `lambda_(e)`decrease with increase in `phi and lambda_(ph)`B. `lambda_(e)` is approximentily balved , if d is doubledC. for large potential diffrence `(Vgtgt phi //e). Lambda_(e) `is approximately halved if V is made four timeD. `lambda_(e) ` increase at the same rate as `lambda_(ph) for lambda_(ph) lt hc//phi`
Answer» Correct Answer - C The wavelength of emitted photoelectron as per de brogher is ` lambda_(0) = (h)/(p) = (h)/(sqrt(2m (K.E))` when `lambda_(ph) increases N_(ph) decreases ` K.E. decreases and therefore I,e increse let is independent of the decreases d `Also(hc)/(lambda_(ph)) + eV - phi = (h^(2))/(2m lambda_(00)^(2)) [lambda = (h)/(sqrt(2mK . E))]` `:. (hc)/(e lambda_(ph)) + V (ph)/(e) = (b^(2))/(2me lambda_(0)^(2)))` ....(i) `for V gtgt (phi)/(e) , phi ltlt eV` ` also (hc)/(e lambda_(ph)) ltlt V ` Then from eq (i) `lambda_(e) prop (1)/(sqrt(V))` Therefore if V is mode our times `lambda_(e)` is approximated half

Discussion

368.	An electron in nth excited state in a hydrogen atom comes down to first excited state by emitting ten different wavelength. Find value of `n` (an integer).A. 6B. 7C. 8D. 9
Answer» Correct Answer - A

Discussion

369.	The potential different across the Coolidge tube is `20 kV and 10 m A` current flows through the voltage supply. Only `0.5%` of the energy carried by the electrons striking the largest is converted into X-ray. The power carried by the X-ray beam is `p`. ThenA. `0.1W`B. `1W`C. `2W`D. `10W`
Answer» Correct Answer - b

Discussion

370.	In a common base ampifier , the phase difference between the input signal and output voltage isA. ` pi`B. `(pi)/(4)`C. `(pi)/(2)`D. ``0`
Answer» Correct Answer - D Zero , in common base amplifler circuit , input and output vollage are in the plase

Discussion

371.	A `280 `days old cative substance shown an activity of `6000 dps 100 days` later its activity between `3000` days what was its initial activity ?A. `20000 dps`B. `24000 dps`C. `12000 dps`D. `6000 dps`
Answer» Correct Answer - B In two half lives, the activity , will remain `(1)/(4) ` of its initial activity

Discussion

372.	A small particle of mass m moves in such a way that the potential energy `U = ar^2`, where a is constant and r is the distance of the particle from the origin. Assuming Bhor model of quantization of angular momentum and circular orbits, find the rodius of nth allowed orbit.
Answer» Correct Answer - A::B::D The force at a distance r is` `F =- (dU)/(dr) =- 2ar` Suppose r be the radius of nth orbit. Then, the necessary centripetal force is porvided by the above force. Thus, `(mv^2)/(r ) =2ar … (i) ` Further, the quantization of angular momentum gives` `mvr = (nh)/(2pi) ...(ii)` Solvig Eqs. (i) and (ii) for r, we get` `r = ((n^2h^2)/(8am pi^2))^(1//4).

Discussion

373.	Uranium ores on the earth at the present time typically have a composition consisting of 99.3% of the isotope `_92U^238` and 0.7% of the isotope `_92U^235`. The half-lives of these isotopes are `4.47xx10^9yr` and `7.04xx10^8yr`, respectively. If these isotopes were equally abundant when the earth was formed, estimate the age of the earth.
Answer» Correct Answer - A Let `N_0` be number of atoms of each isotope at the time of formation of the earth `(t=0)` and `N_1` and `N_2` the number of atoms at present `(t=t)`. Then, `N_1=N_0e^(-lambda_1t)` …(i) and `N_2=N_0e^(-lambda_2t)` …(ii) :. `N_1/N_2=e^((lambda_2-lambda_1)t)` ...(iii) Further it is given that `N_1/N_2=99.3/0.7` ...(iv) Equating Eqs. (iii) and (iv) and taking log on both sides, we have `(lambda_2-lambda_1)t=1n(99.3/0.7)` :. `t=(1/(lambda_2-lambda_1))1n(99.3/0.7)` Substituting the values, we have `t=(1)/((0.693)/(7.04xx10^8)-(0.693)/(4.47xx10^9))1n(99.3/0.7)` or `t=5.97xx10^9yr`

Discussion

374.	The element curium `_96^248 Cm` has a mean life of `10^13s`. Its primary decay modes are spontaneous fission and `alpha`-decay, the former with a probability of 8% and the later with a probability of 92%, each fission releases 200 MeV of energy. The masses involved in decay are as follows `_96^248 Cm=248.072220 u`, `_94^244 P_u=244.064100 u` and `_2^4 He=4.002603u`. Calculate the power output from a sample of `10^20` Cm atoms. (`1u=931 MeV//c^2`)
Answer» Correct Answer - A::B::C The reaction involved in `alpha`-decay is `_96^248Cmrarr_94^244Pu+_2^4He` Mass defect, `Deltam=mass of _96^248Cm-mass of _94^244Pu-massof_2^4He` `=(248.072220-244.064100-4.002603)u` `=0.005517u` Therefore, energy released in `alpha`-decay will be `E_alpha=(0.005517xx931)MeV=5.136MeV` Similarly, `E_(fission)=200MeV` (given) Mean life is given as `t_(mean)=10^13s=1//lambda` :. Disintegration constant `lambda=10^-13s^-1` Rate of decay at the moment when number of nuclei are `10^20` `=lambdaN=(10^-13)(10^20)` `=10^7` disintegration per second Of these, 8% are in fission and 92% are in `alpha`-decay. Therefore, energy released per second `=(0.08xx10^7xx200+0.92xx10^7xx5.136)MeV` `=2.074xx10^8MeV` :.Power output (in watt) = energy released per second`(J//s)`. `=(2.074xx10^8)(1.6xx10^-13)` `=3.32xx10^-5Js^-1` :. Power output `=3.32xx10^-5 W`

Discussion

375.	A beam of light has three wavelengths `4144 Å`, and `6216 Å` with a total instensity of `3.6 xx 10^(-3) Wm^(-2)` equally distributed amongst the three wavelengths. The beam falls normally on an area `1.0 cm^2` of a clean metallic surface of work function 2.3 eV. Assume that there is no loss of light by reflection number of photoelectrons liberated in two seconds.
Answer» Correct Answer - A::B Energy of photon having wavelength `4144 Å`, `E_1 = (12375)/(4144) eV` `=2.99 eV` Similarly, `E_2 = (12375)/(4972) eV` = 2.49 eV and `E_3 (12375)/(6216) eV` =1.99 eV Since, only `E_1` and `E_2` are greater than the work function `W = 2.3 ev`, only first two wavelengths are capable for ejecting photoelectrons. Given intensity is equally distributed in all wavelengths. Therefore, instensity corresponding to each wavelength is `(3.6xx10^(-3))/(3) = 1.2 xx10^(-3) W/m^2` Or energy incident per second in the given area (A = 1.0 cm^2 = 10^(-4) m^2)` is `P=1.2xx10^(-3)xx10^(-4)` `= 1.2 xx10^(-7) J/s` Let `n_1` be the number of photons incident per unit time in the ginven area corresponding to first wavelength. Then, `n_1 = (p)/(E_1)` `= (1.2xx10^(-7)/(2.99xx1.6xx10^(-19)` `=2.5xx10^11` Similarly, `n_2=(P)/(E_2)` `=1.2xx10^(-7)/(2.49xx1.6xx10^(-19)` `=3.0xx10^11` Since each energetically capable photon ejects one electron, total number of photoelectrons liberated in 2 s.

Discussion

376.	Electrons with de - Brogli wavelengtyh `lambda` fall on the target in an X-ray tube.The cut off wavelength of the emitted Xrays is (a) `lambda_0 = (2mclambda^2)/(h)` (b)`lambda_0 = (2h)/(mc)` (c ) `lambda_0 (2m^2 c^2 lambda^3)/(h^2)` (d)`lambda_0 = lambda`
Answer» Correct Answer - A::B::C::D Momentum of bombarding electrons,` `p = (h)/(lambda)` `:. Kinetic energy of bombarding electrons, `K = (p^2)/(2m) = (h^2)/(2mlambda^2)` This is also maximum energy of X-ray photons. Therefore, `(hc)/(lambda_0) = (h^2)/(2mlambda^2) or `lambda_0 = (2mlambda^2 c)/(h)` :. Correct option is (a).

Discussion

377.	Determine the energy of the characteristic X-ray `(K_beta)` emitted from a tungsten (Z = 74) target when an electron drops from the M-shell (n=3) to a vacancy in the K-shell (n=1)
Answer» Correct Answer - B::C::D Energy associated with the electron in the K-shell is approximately `E_K =- (74-1)^2 (13.6 eV) =- 72474 eV` An electron in the M-shell is subjected to an effective nuclear that depends on the number of electrons in the n=1 and n =2 states because these electrons shield the M electrons form the nucleus. Because there are eight electrons in the n=2 state and one remaining in the n=1 state, roughly nine electrons shield M electrons from the nucleus. So, `Z_(eff) = z-9` Hence, the energy associated with an electron in the M-shell is ` E_M = (-13.6Z_(eff)^2)/(3^2) eV = (-13.6(Z- 9)^2)/(3^2) eV` `=(13.6)(74-9)^2/(9) eV =- 6384 eV` Therefore, emitted X-ray has an energy equal to `E_M - E_K = {-6384 - (-72474)}eV =- 66090eV`

Discussion

378.	The currect voltage relation of a diode is given by `1 = (e^(van v//T) -1 )mA` where the applied volied `V` is in volts and the tempetature `T` is in degree kelvin if a student make an error meassurting `+- 01 V` while measuring the current of `5 mA at 300 K` what be the error in the value of current in `mA`A. `0.2 mA`B. `0.02 mA`C. `0.5 mA`D. `0.05 mA`
Answer» Correct Answer - A The current voltage relat5ion of diode is `1 = (e^(1000V//T) - 1) mA `(given) when 1 = 5 mA, (e^(1000V//T)= 6 mA` `Also dl = (e^(1000V//T) xx (1000)/(T) (By exponential function) `= ( 6 mA) xx (1000)/(300) xx (0.01)` `= 0.2mA`

Discussion

379.	In an X- ray tube , electrons accelerated through a potential different of `15000` volts strike a copper target . The speed of the emitted X - ray inside the tube is …….. `m//s`
Answer» Correct Answer - A::C The speed of X - rays is always `3 xx 10^(8) m//s ` in vacuum . Which electrons are acceleated in are X - rays tube .

Discussion

380.	To produce characteristic `X` - rays using a Tungsten target in an x - ray generator , the accelerating should be greater then ……. Volts and the energy of the characteristion is …… eV . (The binding energy of the intermost electron in Tungsten is - 40keV).
Answer» Correct Answer - C For minimum accelerating voltage , the electron should jump from `n = 2`to`n = 1leval` for characterintic X - rays `(1)/(lambda) = R_(a)(L-1)^(2) [1 - (1)/(n^(2))] = (E)/ (hc) ( E_(1))/(hc) = R_(a) (Z - 1)^(2)[1 -(1)/(2^(2)) ] ` …..(i) The binding energy of innermost eletron ` = 40 keV` ` `:.` lonisation potential of tungsten ` = 40 kV = 40 xx 10^(2) V` `rArr (E_(2))/(hc) = R_(a) (Z - 1)^(2)[1 -(1)/(2^(2)) ] ` ....(ii) ` (E_(1))/(E_(2)) = [[1 - (1)/(2^(2))]]/([1 - (1)/(oo^(2)]]` `rArr E_(1) = (3)/(4) E_(2) = (3)/(4) xx 40000 eV = 30000 eV` `:. ` Minimum accelerating vollage `V_(min) = (E_(1))/(e) = 30000 V `

Discussion

381.	A triod has plate characteristies in the from of purallet lines in the region of our interest At a given in terms of platevoltage V (in volts ) by the algebraic relation `1 = 0.125V - 7.5` for grid of - 3 volts is 5 millanpers , determine the plate resistance of(r_(p)) transcondutance (g)and the arrplfication facter (u) for the trid
Answer» Correct Answer - A::B::C `1= 0.125V - 7.5` rArr dl = 0.125dV or (dV)/(dl) = (1)/(0.125) = 8` we know that plate resistance, r_(p) = (dV)/(dl) = 8m Omega` The transconductance, `g_(m)= [(dl)/(dV_(g))]_(V= conslt)` `At V_(g) = -1 volt , V = 300 volt,` the plate current l = [0.125 xx 300 - 7.5 ] mA = 30 mA` Also is given that ` V_(g) = - 3V , V = 300V and l = 5 mA` `:. g_(m) = [(30 - 5)/(- 1 -(1 -3))] = (25)/(2) xx 10^(-3) = 12.5 xx 10^(-3) s ` The characteristics are given in the from of purallel lines Amplification factor `= r_(p)g_(m) = 8 xx 10^(3) xx 12.5 xx 10^(-3) = 100`

Discussion

382.	A double ionised lithium atom is hydrogen like with atomic number `3` (i)Find the wavelength of the radiation to excite the electron in `Li^(++)`from the first to the third bohr orbit (lonisection energy of the hydrogen atom equals `13. 6)` (ii) How many spectral lines are observed in the emission spetrum of the above excled system ?
Answer» Correct Answer - A::B::C::D (i)` E_(0) = (13.6)/(n^(2))Z^(2) eV //atom ` for `Li^(2+ ), Z = 3 :. E_(n) = (-13.6 xx 9)/(n^(2)) eV atom` `:. E_(1) = (13.6 xx 9)/(1) and E_(3) = -(13.6 xx 9)/(9) = - 13.6` ` deltaE = E_(3)-E_(1) = - 13.6 - (-13.6 xx 9)` `= 13.6 xx 8 = 108.8eV// atom ` `lambda= (12400)/(E(in eV)) Å= (12400)/(108.8) = 114Å`

Discussion

383.	A cesium photocell, with a steady potential difference of 60 V across it ,is illuminated by a small bright light placed 1m away. When the same light is placed 2m away, the electrons crossing the photocellA. Each carry one quarter of their previous momentumB. Each carry one quarter of their previous energyC. Are one quarter as numerousD. Are half as numerous
Answer» Correct Answer - C

Discussion

384.	The electric field applied in vertical directional to the cathode rays moving horizontally deflect them in __________A. horizontal directionB. vertical directionC. Both (a) and (b)D. None of these
Answer» Correct Answer - B

Discussion

385.	It is proposed to use the nuclear fusion reaction, `_1^2H+_1^2Hrarr_2^4He` in a nuclear reactor 200 MW rating. If the energy from the above reaction is used with a 25 per cent efficiency in the reactor, how many grams of deuterium fuel will be needed per day?(The masses of `_1^2H` and `_2^4He are 2.0141 atommic mass units and 4.0026 atomic mass units respectively.)
Answer» Correct Answer - A::B Mass defect in the given nuclear reaction, `Deltam=2(mass of deuterium)-(mass of helium)` `=2(2.0141)-(4.0026)=0.0256` Therefore, energy released `DeltaE=(Deltam)(931.48)MeV=23.85MeV` `=23.85xx1.6xx10^-13J=3.82xx10^-12J` Efficiency is only 25%, therefore, `25% of DeltaE=(25/100)(3.82xx10^-12)J` `=9.55xx10^-13J` i.e. by the fusion of two deuterium nuclei, `9.55xx10^-13J` energy is available to the nuclear reactor. Total energy required in one day to run the reactor with a given power of 200 MW, `E_(Total)=200xx10^6xx24xx3600=1.728xx10^13J` :. Total number of deuterium nuclei required for this purpose, `n=(E_(Total))/(DeltaE//2)=(2xx1.728xx10^13)/(9.55xx10^-13)` `=0.362xx10^26` :. `Mass of deuterium required = (Number of g-moles of deuterium required)xx2g` `=((0.362xx10^26)/(6.02xx10^23))xx2=120.26g`

Discussion

386.	Calculate the Q-values of the following fusion reactions:(a)`1^2H+1^2H rarr 1^3H+1^1H`.`1^2H+1^2H rarr 2^3(He)+n` `1^2H+1^3H rarr 2^4(He)+n`.Atomic masses are `m(1^2H)=2.014102 u`, `m(1^3H)=3.016049 u`, `m(2^3(He))=3.016029 u`, `m(2^4(He))=4.002603 u`.
Answer» Correct Answer - A::B::C::D Q-value`=(Deltam)(931.5)MeV` (a) Q-value `=(2xx2.014102-3.016049-1.007825)xx931.5` `=4.05 MeV` Similarly, Q-value of other parts can also be obtained.

Discussion

387.	One of the lines in the emission spectrum of `Li^(2 +)` has the same wavelength as that of the second line of Balmer series in hydrogen spectrum. The electronic transition correspnding to this line is.A. `12to9`B. `12to6`C. `4to3`D. `6to3`
Answer» Correct Answer - B

Discussion

388.	The photons from the balmer series in Hydrogen spectrum having wavelength between `450 nm` to `700 nm` are incident on a metal surface of work function `2 eV` find the maximum kinetic energy os ejected electron (Given hc = 1242 eV nm)`
Answer» KEY CONCEPT : The wavelength `lambda` of photo for different lines of bamber series is given by `(hc)/(lambda)= 13.6 [(1)/(2^(2)) - (1)/(n^(2))] eV where n = 3,4,5` Using above relation , we get the value of `lambda = 657 nm , 487 nm between 450nm and 700 nm ` is smaller then `657 nm` , electron of max K.E will be emitted for photon corresponding to wavelength `487 nm` with `(K.E) = (hc)/(lambda) - W = ((1242)/(487) - 2) = 0.55 eV`

Discussion

389.	The photons from the Balmer series in Hydrogen spectrum having wavelength between `450 nm` to `700 nm` are incident on a metal surface of work function `2 eV` Find the maximum kinetic energy of ejected electron (Given hc = 1242 eV nm)`
Answer» Balmer series `lambda_(32) = (12375)/(E_3 - E_2) = (12375)/((13.6)((1)/(4)- (1)/(9))` =6551 Å = 655.1 nm `lambda_(42) = (12375)/(E_4 -E_2) = (12375)/((13.6)((1)/(4) - (1)/(16)))` = 4853 Å =433.3 nm First two lie in the given range. of these `lambda_(42)` corresponds to more energy `E = E_4 - E_2 = (13.6)((1)/(4) - (1)/(16))` =2.55 eV `K_(max) = E -W = (2.55 - 2.0) eV` = 0.55 eV.

Discussion

390.	A 1000 W transmitter works at a frequency of 880kHz. The number of photons emitted per second IsA. `1.7xx10^28`B. `1.7xx10^30`C. `1.7xx10^23`D. `1.xx10^25`
Answer» Correct Answer - B Number of photons emitted per second `=(Energy radiated per second)/(Energy of one photon) = (p)/(hf) `= (1000)/(6.63xx10^(-34)xx880xx10^3)` `1.7xx10^(30)`.

Discussion

391.	Assertion: X-rays cannot be obtained in the emission spectrum of hydrogen atom.Reason: Maximum energy of photons emitted form hydroen spectrum is 13.6 eV.A. If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.B. If both Assertion and Reason or true but Reason is not the correct explanation of Assertion.C. If Assertion is true, but the Reason is false.D. If Assertion is false but the Reason is true.
Answer» Correct Answer - A Energy of x-raygt 13.6 eV

Discussion

392.	When the intensity of a light source is increased (a) the number of photons emitted by the source in unit time increases (b) the total energy of the photons emitted per unit time increases (c) more energetic photons are emitted (d) faster photons are emittedA. `a,b`B. `a,c`C. `a,d`D. `b,d`
Answer» Correct Answer - A

Discussion

393.	Photons with energy `5 eV` are incident on a cathode `C` in a photoelectric cell . The maximum energy of emitted photoelectrons is `2 eV`. When photons of energy `6 eV` are incident on `C` , no photoelectrons will reach the anode `A` , if the stopping potential of `A` relative to `C` isA. 5B. 3 VC. 1 VD. 4 V
Answer» Correct Answer - B

Discussion

394.	Consider the electron energy level diagram of H-atom Photons associated with shortes and longest wavelength would be emitted from the atom by the transitions labeled A. D and C respcitivelyB. C and A respectivelyC. C and D respectivleyD. A and C respectively
Answer» Correct Answer - C

Discussion

395.	Assertion: if wavelength of light is doubled, energy and momentum of photons are reduced to half. Reason: By increasing the wavelength, speed of photons will decrease.A. If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.B. If both Assertion and Reason or true but Reason is not the correct explanation of Assertion.C. If Assertion is true, but the Reason is false.D. If Assertion is false but the Reason is true.
Answer» Correct Answer - C `E = (hc)/(lambda) and P =(h)/(lambda)` `:. E and P prop (1)/(lambda)` Speed of all wavelengths (in vacuum) is c.

Discussion

396.	The maximum kinetic energy of photoelectrons emitted from a surface when photons of energy 6 eV fall on it is 4eV. The stopping potential in volt is (a)2 (b) 4 (c ) 6 (d) 10
Answer» Correct Answer - B Stopping potential is the negative potential applied to stop the electorns having maximum kinetic energy. Therefore, stopping potential will be 4v.

Discussion

397.	In a fall wave rectifer circuit operating from `50 Hz` mains frequency , the fundamental frequency in the ripple would beA. `25 Hz`B. `50 Hz`C. `70.7 Hz`D. `100 Hz`
Answer» Correct Answer - D input frequency `f = 50 Hz rArr T = (1)/(50)` for fall wave rectifer ` T_(1) = (T)/(2) = (1)/(100) rArr f_(1) = 100 Hz`

Discussion

398.	The collector plate in an experiment of photoelectric effect is kept vertical above the emitter plate. Light source is put on, and a saturation photocurrent is recorded. An electric field is switched on which has a vertically downward direction-A. Photocurrent will increase.B. Kinetic energy of electrons will increaseC. Stopping potential will decreaseD. threshold wavelength increase
Answer» Correct Answer - B

Discussion

399.	An electron is accelerated by a potential difference of 25 volt. Find the de- Broglie wavelength associated with it.
Answer» Correct Answer - B For an electron, de Brogli wavelength is given by. `lambda = sqrt(150)/(V) = sqrt(150)/(25)` `=sqrt6~~2.5Å.`

Discussion

400.	An a- particle and a proton are accelerated from rest by a potential difference of 100V. After this, their de-Broglie wavelengths are `lambda_a` and `lambda_p` respectively. The ratio `(lambda_p)/(lambda_a)`, to the nearest integer, is.
Answer» Correct Answer - A::C `:. lambda =(h)/(p) = (h)/(sqrt2qVm or `lambda prop(1)/(sqrtqm` `(lambda_p)/lambda_a) = sqrt((qa)/(qp). (m_a)/(m_p))` = `sqrt((2)(4))/((1)(1)) =2.828` . The nearest integer is 3. :.Answer is 3.

Discussion

Explore topic-wise InterviewSolutions in .

Hydrogen atom is exited from ground state to another state with principal quantum number equal to `4` Then the number of spectral linear in the emission spectra will beA. `2`B. `3`C. `5`D. `6`

de-Broglie wavelength of an electron in the nth Bohr orbit is `lambda_(n)` and the angular momentum is `J_(n)` thenA. `J_(n)proplambda_(n)`B. `lambda_(n)prop(1)/(J_(n))`C. `lambda_(n)propJ_(n)^(2)`D. None of these

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A silver balll is suspended by a string in a vacuum chamber and ultraviolet light of wavelength `2000 Å` is directed at it. What electrical potential will the ball acquire as a result? Work function of silver is 4.3 eV.

The short-wavelength limit shifts by 26 pm when the operating voltage in an X-ray tube is increased to 1.5 times the original value. What was the original value of the operating voltage?

if `lambda_(Cu)` is the wavelength of `K_alpha`, X-ray line fo copper (atomic number 29) and `lambda_(Mo)` is the wavelength of the `K_alpha` X-ray line of molybdenum (atomic number 42), then the ratio `lambda_(Cu)/lambda_(Mo)`is close to (a) 1.99 (b) 2.14 (c ) 0.50 (d) 0.48

Wavelength of `K_alpha` line of an element is `lambda_0`. Find wavelength of `K_beta` - line for the same elemetn.

From what meterial is the anod of an X-ray tube made if the `K_alpha` line wavelength of thej characteristic spectrum is `0.76 Å`?

The threshold wavelength for photoelectric from a material is `5200 Å` photoelectric will be emittillumnated with when this meterial is illiuminated with monecharnatic radiation from aA. `50 `watt infrared lampB. `1` - watt infra- red lampC. `50 `watt ubraviolet lampD. `1` - watt ahraviolet lamp

An electron in nth excited state in a hydrogen atom comes down to first excited state by emitting ten different wavelength. Find value of `n` (an integer).A. 6B. 7C. 8D. 9

The potential different across the Coolidge tube is `20 kV and 10 m A` current flows through the voltage supply. Only `0.5%` of the energy carried by the electrons striking the largest is converted into X-ray. The power carried by the X-ray beam is `p`. ThenA. `0.1W`B. `1W`C. `2W`D. `10W`

In a common base ampifier , the phase difference between the input signal and output voltage isA. ` pi`B. `(pi)/(4)`C. `(pi)/(2)`D. ``0`

A `280 `days old cative substance shown an activity of `6000 dps 100 days` later its activity between `3000` days what was its initial activity ?A. `20000 dps`B. `24000 dps`C. `12000 dps`D. `6000 dps`

A small particle of mass m moves in such a way that the potential energy `U = ar^2`, where a is constant and r is the distance of the particle from the origin. Assuming Bhor model of quantization of angular momentum and circular orbits, find the rodius of nth allowed orbit.

Electrons with de - Brogli wavelengtyh `lambda` fall on the target in an X-ray tube.The cut off wavelength of the emitted Xrays is (a) `lambda_0 = (2mclambda^2)/(h)` (b)`lambda_0 = (2h)/(mc)` (c ) `lambda_0 (2m^2 c^2 lambda^3)/(h^2)` (d)`lambda_0 = lambda`

Determine the energy of the characteristic X-ray `(K_beta)` emitted from a tungsten (Z = 74) target when an electron drops from the M-shell (n=3) to a vacancy in the K-shell (n=1)

In an X- ray tube , electrons accelerated through a potential different of `15000` volts strike a copper target . The speed of the emitted X - ray inside the tube is …….. `m//s`

To produce characteristic `X` - rays using a Tungsten target in an x - ray generator , the accelerating should be greater then ……. Volts and the energy of the characteristion is …… eV . (The binding energy of the intermost electron in Tungsten is - 40keV).

The electric field applied in vertical directional to the cathode rays moving horizontally deflect them in __________A. horizontal directionB. vertical directionC. Both (a) and (b)D. None of these

Calculate the Q-values of the following fusion reactions:(a)`1^2H+1^2H rarr 1^3H+1^1H`.`1^2H+1^2H rarr 2^3(He)+n` `1^2H+1^3H rarr 2^4(He)+n`.Atomic masses are `m(1^2H)=2.014102 u`, `m(1^3H)=3.016049 u`, `m(2^3(He))=3.016029 u`, `m(2^4(He))=4.002603 u`.

One of the lines in the emission spectrum of `Li^(2 +)` has the same wavelength as that of the second line of Balmer series in hydrogen spectrum. The electronic transition correspnding to this line is.A. `12to9`B. `12to6`C. `4to3`D. `6to3`

The photons from the balmer series in Hydrogen spectrum having wavelength between `450 nm` to `700 nm` are incident on a metal surface of work function `2 eV` find the maximum kinetic energy os ejected electron (Given hc = 1242 eV nm)`

The photons from the Balmer series in Hydrogen spectrum having wavelength between `450 nm` to `700 nm` are incident on a metal surface of work function `2 eV` Find the maximum kinetic energy of ejected electron (Given hc = 1242 eV nm)`

A 1000 W transmitter works at a frequency of 880kHz. The number of photons emitted per second IsA. `1.7xx10^28`B. `1.7xx10^30`C. `1.7xx10^23`D. `1.xx10^25`

When the intensity of a light source is increased (a) the number of photons emitted by the source in unit time increases (b) the total energy of the photons emitted per unit time increases (c) more energetic photons are emitted (d) faster photons are emittedA. `a,b`B. `a,c`C. `a,d`D. `b,d`

Consider the electron energy level diagram of H-atom Photons associated with shortes and longest wavelength would be emitted from the atom by the transitions labeled A. D and C respcitivelyB. C and A respectivelyC. C and D respectivleyD. A and C respectively

The maximum kinetic energy of photoelectrons emitted from a surface when photons of energy 6 eV fall on it is 4eV. The stopping potential in volt is (a)2 (b) 4 (c ) 6 (d) 10

In a fall wave rectifer circuit operating from `50 Hz` mains frequency , the fundamental frequency in the ripple would beA. `25 Hz`B. `50 Hz`C. `70.7 Hz`D. `100 Hz`

An electron is accelerated by a potential difference of 25 volt. Find the de- Broglie wavelength associated with it.

An a- particle and a proton are accelerated from rest by a potential difference of 100V. After this, their de-Broglie wavelengths are `lambda_a` and `lambda_p` respectively. The ratio `(lambda_p)/(lambda_a)`, to the nearest integer, is.