394 + Interview Questions in DUAL-NATURE-OF-RADIATION-AND-MATTER IN GENERAL AWARENESS Page 1 InterviewSolution

1.	The wavelength of a photon needed to remove a proton from a nucleus which is bound to the nucleus with 1MeV energy is nearlyA. `1.2nm`B. `1.2xx10^(-3)nm`C. `1.2xx10^(-6) nm`D. `.12xx10^(1)nm`
Answer» Correct Answer - b Energy `E=1MeV=10^(6)eV , hc=1240 eV nm` `E=(hc)/lambda or lambda=(hc)/E=(1240 eVnm)/(10^(6)eV)=1.24xx10^(-3)nm`

Discussion

2.	Energy of `K` - shell electron be `-40000 eV`. If `60000 V` potential is applied at Coolidge tube then which of the following `X` - rays will get form ?A. ContinuousB. White `X` - raysC. Continuous and all series of characteristicD. None of these
Answer» Correct Answer - C When applied voltage is greater then energy of `K` - electron , continuous and all characteristics `X` - rays are emitted.

Discussion

3.	The wavelength of a photon needed to remove a proton from a nucleus which is bound to the nucleus with 1MeV energy is nearlyA. 1.2 nmB. `1.2xx10^(-3)` nmC. `1.2xx10^(-6)`nmD. `1.2xx10` nm
Answer» Given in the question, Energy of photon, `E=1MeVimplies=10^(6)eV` Now, `" "hc=1240 eVnm` Now, `" "E=(hc)/(lamda)` `implies" "l=(hc)/(E)=(1240eVnm)/(10^(6)eV)` `" "=1.24xx10^(-3)` nm

Discussion

4.	What determines the hardness of the `X` - rays obtained from the Coolidge filament ?A. Current in the filamentB. Pressure of air in the tubeC. Nature of targetD. Potential difference between cathode and target
Answer» Correct Answer - D Hard `X `- rays are of higher energy and the energy of `X` - rays depends on the potential difference between the cathode and the target.

Discussion

5.	It is harder to remove free electron from copper than from sodium. Which has higher work function?
Answer» The work function of copper is higher than that of sodium.

Discussion

6.	What determines the maximum velocity of the photoelectrons?
Answer» The frequency of the incident radiations and work function of the metallic surface.

Discussion

7.	A photon of wavelength `6630 Å` is incident on a totally reflecting surface . The momentum delivered by the photon is equal toA. `6.63 xx 10^(-27) kg - m//sec`B. `2 xx 10^(-27) kg - m//sec`C. `10^(-27) kg - m//sec`D. None of these
Answer» Correct Answer - B The momentum of the incident radiation is given as `p = (h)/(lambda)`. When the light is totally reflected normal to the surface the direction of the ray is reversed . That means it reverses the direction of its momentum without changing its magnitude `:. Delta p = 2 p = ( 2h)/(lambda) = ( 2 xx 6.6 xx 10^(-34))/(6630 xx 10^(-10))` ` = 2 xx 10^(-27) kg - m // sec`.

Discussion

8.	The work function of copper is 4.0 eV. If two photons, each of energy 2.5 eV strike with some electrons of copper, will the emission be possible?
Answer» One photon can eject one photoelectron from the surface of a metal provided its energy is not less than the work function of metal. Since each photon has energy less than the work function of copper, hence, photoelectric emission is not possible.

Discussion

9.	An electron and a photon have same wavelength `lambda`, what is ratio of their kinetic energies ?A. `h/(m_eclambda)`B. `(2m_e clambda)/h`C. `1:1`D. `h/(2m_eclambda)`
Answer» Correct Answer - 4

Discussion

10.	The number of photons `(lambda = 6630 Å)` that strike per second on a totoally reflecting screen (as shown in figure), so that a force of `1N` is exerted on the screen, is approximentely). A. `10^23`B. `10^27`C. `10^25`D. `10^26`
Answer» Correct Answer - 2

Discussion

11.	If the cathode - anode potential difference in an `X` - ray tube be `10^(5) V` , then the maximum energy of `X` - ray photon can beA. `10^(5) J`B. `10^(5) MeV`C. `10^(-1) MeV`D. `10^(5) KeV`
Answer» Correct Answer - C Since `lambda_(min) = (12375)/(V) Å = (12375)/(10^(5)) Å = 0.123 Å` `E_(max) = ( hc) /(lambda_(min))`, On putting the values . `E_(max) ~= 10^(-1) MeV`.

Discussion

12.	(a) Estimate the speed with which electrons emitted from a heated cathode of an evacuated tube impinge on the anode maintained at a potential difference of 500 V with respect to the cathode. Ignore the small initial speeds of the electrons. The specific charge to of the electron, i.e., its `e//m` is given to `1.76xx10^(11)Ckg^(-1)`. (b) Use the same formula you employ in (a) to obtain electron speed for an anode potential of 10 MV. Do you see what is wrong? In what way is the formula to be modified?
Answer» (a)Potential difference across the evacuted tube , V=500 V Specific charge of an electron, e/m=`1.76xx10^(11) C kg^(-1)` The speed of each emitted electron is given by the relation for kinetic energy as : `KE=1/2mv^2=eV` `therefore v=((2eV)/m)^(1/2)=(2Vxxe/m)^(1/2)` `=(2xx500xx1.76xx10^11)^(1/2)=1.327xx10^(7)` m/s Therefore, the speed of each emitted electron is `1.327xx10^7` m/s (b)Potential of the anode , V=10 MV =`10xx10^6` V The speed of each electron is given as : `v=(2Ve/m)^(1/2)` `=(2xx10^7 xx 1.76 xx 10^11)^(1/2)` `=1.88xx10^(9)` m/s This result is wrong because nothing can move faster than light. In the above formula, the expression `(mv^2//2)` for energy can only be used in the non-relativistic limit, i.e. , for v lt lt c. For very high speed problems , relativistic equations must be considered for solving them.In the relativistic limit , the total energy is given as : `E=mc^2` Where, m=Relativistic mass `=m_0(1-v^2/c^2)^(1/2)` `m_0` = Mass of the particle at rest Kinetic energy is given as : `K=mc^2-m_0c^2`

Discussion

13.	If the potential difference between the anode and cathode of the `X` - ray tube is increases A. The peaks at `R` and `S` would move to shorter wavelengthB. The peaks at `R` and `S` would remain at the same wavelengthC. The cut off wavelength at `P` would decreaseD. `(b)` and `( c)` both are correct
Answer» Correct Answer - D Peaks on the graph represent characteristic `X` - ray spectrum. Every peak has a certain wavelength , which depends upon the transition of electrons inside the atom of the target. While `lambda_(min)` depends upon the accelerating voltage (As . `lambda_(min) prop 1//V`).

Discussion

14.	Calculate the energy of an electron having de-Broglie wavelength 5500Å. Given, `h=6.62xx10^(-34)Js, m_(e)=9.1xx10^(-31)kg`.
Answer» Correct Answer - `4.9xx10^(-6)eV` Here, `lambda=5500Å =5500xx10^(-10)m` `55xx10^(-8)m` de-Broglie wavelength associated with electron is `lambda=h/(sqrt(2m_(e)E)) or E=(h^(2))/(2m_(e)lambda^(2))=1/(2m_(e)) (h/lambda)^(2)` `:. E=1/(2xx(9.1xx10^(-31)))[(6.6xx10^(-34))/(55xx10^(-8))]^(2)` `=7.9xx10^(-25)J` `=(7.9xx10^(-25))/(1.6xx10^(-19))=4.9xx10^(-6)eV`

Discussion

15.	Find the (a) maximum frequency and (b) minimum wave-length of X-rays produced by 30 kV electrons. Given, `h=6.63xx10^(-34)Js`.
Answer» Correct Answer - A (a) : Here, `V=kV,h=6.63xx10^(-34)Js` As `eV=hv_(max)orv_(max)=(eV)/(h)=(1.6xx10^(-19)xx3xx10^(4))/(6.63xx10^(-34))` `=7.24xx10^(18)Hz` As `eV=(hc)/(lamda_(min))orlamda_(min)=(hc)/(eV)=(6.63xx10^(-34)xx3xx10^(8))/(1.6xx10^(-19)xx3xx10^(4))` `=0.041xx10^(-9)m=0.041nm`

Discussion

16.	The kinetic energy of an electron is `5 eV`. Calculate the de - Broglie wavelength associated with it `( h = 6.6 xx 10^(-34) Js , m_(e) = 9.1 xx 10^(-31) kg)`A. `5.47 Å`B. `109 Å`C. `2.7 Å`D. None of these
Answer» Correct Answer - A `lambda = (h)/(sqrt(2 m E)) = (6.6 xx 10^(-34))/(sqrt(2 xx 9.1 xx 10^(-31) xx 5 xx 1.6 xx 10^(-19))` `= 5.469 xx 10^(-10) m = 5.47 Å`

Discussion

17.	Calculate the (a) momentum and (b) de-Broglie wavelength of the electrons accelerated through a potential difference of 56V. Given, `h=6.63xx10^(-34)Js, m_(e)=9.1xx10^(-31)kg, e=1.6xx10^(-19)C`.
Answer» (a) `eV=1/2mv^(2) or mv = sqrt(2eVm)` Momentum of electron, `mv=sqrt(2xx(1.6xx10^(-19))xx56xx9xx10^(-31))=4.02xx10^(-24)kgms^(-1)` (b) `lambda=12.27/(sqrtV)Å=12.27/(sqrt(56))=1.64Å`

Discussion

18.	What is the (a) momentum (b) speed and (c) de-Broglie wavelength of an electron with kinetic energy of 120 eV. Given `h=6.6xx10^(-34)Js, m_(e)=9xx10^(-31)kg , 1eV=1.6xx10^(-19)J`.
Answer» (a) `p=sqrt(2mK)=sqrt(2xx(9xx10^(-31))xx(120xx1.6xx10^(-19)))=5.88xx10^(-24)kgms^(-1)` (b) `v=sqrt((2K)/m)=sqrt((2xx(120xx1.6xx10^(-19)))/(9xx10^(-31)))=6.53xx10^(6)ms^(-1)` (c) `lambda=h/p=(6.63xx10^(-34))/(5.88xx10^(-24))=1.13xx10^(-10)m=1.13Å`

Discussion

19.	The equivalent wavelength of a moving electron has the same value as that of a photon having an energy of `6xx10^(-17)J`. Calculate the momentum of the electron.
Answer» Energy of photon, `E=6xx10^(-17)J` `E=hv=(hc)/lambda` or wavelength of photon, `lambda=(hc)/E` Momentum of moving electron `p=h/lambda=h/(hc//E)=E/c=(6xx10^(-17))/(3xx10^8)` `=2xx10^(-25)kgms^-1`

Discussion

20.	A particle which has zero rest mass and non - zero energy and momentum must travel with a speedA. Equal to c, the speed of light in vacuumB. Tending to infinityC. Greater than cD. Less than c
Answer» Correct Answer - 1

Discussion

21.	The momentum of a photon having energy equal to the rest energy of an electron isA. ZeroB. `2.73xx10^(-22) "kg ms"^(-1)`C. `1.99xx10^(-24) "kg ms"^(-1)`D. Infinite
Answer» Correct Answer - 2

Discussion

22.	What is the momentum
Answer» Given, `KE = 120 eV, m = 9.1xx10^(-3)kg, e = 1.6xx10^(-19)c` `P = sqrt(2m(KE)) = sqrt(2xx9.1xx10^(-31)xx(120xx1.6xx10^(-19))) therefore P = 5.91xx10^(-24)kg-m//s`

Discussion

23.	What is the Speed
Answer» Given, `KE = 120 eV, m = 9.1xx10^(-3)kg, e = 1.6xx10^(-19)c` `upsilon = (p)/(m) = (5.91xx10^(-24))/(9.1xx10^(-31))=6.5xx10^(6)m//s`

Discussion

24.	Assertion : The de - Broglie wavelength of a molecule varies inversely as the square root of temperature. Reason : The root mean square velocity of the molecule depends on the temperature.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If assertion and reason both are false.
Answer» Correct Answer - A de - Broglie wavelength associated with gas molecules varies as `lambda prop (1)/(sqrt(T))`

Discussion

25.	Monochromatic light of frequency `6.0 xx 10^(14) Hz` is produced by a laser. The power emitted is `2 xx 10^(-3)` w. The number of photons emitted, on the average, by the sources per second is
Answer» (a) Each photon has an energy `E = h ν = ( 6.63 xx10^(–34) J s) (6.0 xx10^(14) Hz) = 3.98 xx 10^(–19) J ` (b) If N is the number of photons emitted by the source per second, the power P transmitted in the beam equals N times the energy per photon E, so that P = N E. Then `N=P/E=(2.0xx10^(-3)W)/(3.98xx10^(-19)J)` `= 5.0 xx10^(15)` photons per second.

Discussion

26.	Monochromatic light of frequency `6.0xx10^(14)` Hz is produced by a laser. The power emitted is `2.0 xx10^(-3)` W(a) What is the energy of a photon in the light beam ? (b ) How many photons per second , on an average , are emitted by the source ?
Answer» Each photon has an energy ` E = hv = (6.63xx 10^(-34) J s)(6.0 xx10^(14) Hz)` ` = 3.98 xx 10^(-19)` J If N is the number of photons emitted by the source per second , the power P transmitted in the beam equals N times the energy per photon E, so that P = N E Then N ` P/E = (2.0 xx 10^(-3)W)/(3.98 xx 10^(-19)J)` ` = 5.0 xx10^(15)` photons per second

Discussion

27.	What is the de Broglie wavelength of A dust particle of mass `1.0xx10^(-9)` kg drifting with a speed of 2.2 m/s ?
Answer» Given, for a dust particle `m = 1xx10^(-9)` kg and `upsilon = 2.2 m//s` ` lambda = h/(mv) = (6.63xx 10^(-34))/(1xx10^(-9)xx2.2 ) = 3.0 xx 10^(-25)` m

Discussion

28.	An electron and a photon each has a wavelength of 1.0nm. Find (a) their momenta, (b) the energy of the photon , and © the kinetic energy electron. Take `h=6.63xx10^(-34) Js`.
Answer» (a) For electron or photon, momentum, `p=h/lambda=(6.63xx10^(-34))/(10^(-9))=6.63xx10^(-25)m` (b) `E=(hc)/lambda=((6.63xx10^(-34))xx(3xx10^(8)))/(10^(-9)xx(1.6xx10^(-19)))=1243eV , (c) K=1/2 (p^(2))/m =1/2xx((6.63xx10^(-25))^(2))/(9xx10^(-31)xx(1.6xx10^(-19)))=1.52eV`

Discussion

29.	An electron and a photon each has a wavelength of 1.0nm. Find (a) their momenta, (b) the energy of the photon , and © the kinetic energy electron. Take `h=6.63xx10^(-34) Js`.A.B.C.D.
Answer» Correct Answer - `(i) 6.6xx10^(-25)kgms^(-1), 6.6xx10^(-25)kgms^(-1) (ii) 1.98xx10^(-16) J (iii) 2.39xx10^(-19)J` (i) Momentum of photon. `p=h/lambda=(6.6xx10^(-34))/(1xx10^(-19))` `=6.6xx10^(-25)kgms^(-1)` Momentum of electron, `p=(6.6xx10^(-34))/(1xx10^(-9))` `=6.6xx10^(-25)kgms^(-1)` (ii) Energy of photon, `E=(hc)/lambda` `=(6.63xx10^(-34)xx3xx10^(8))/(1xx10^(-9))=1.98xx10^(-16)J` (iii) Kinetic energy of electron, `K=(p^(2))/(2m)` `=((6.6xx10^(-25))^(2))/(2xx(9.1xx10^(-31)))=2.39xx10^(-19)J`

Discussion

30.	An electron and a photon each have a wavelength 1.00 nm. Find (i) their momenta, (ii) the energy of the photon and (iii) the kinetic energy of electron.
Answer» Given, `lambda = 1 mm = 10^(-9)m, h = 6.63xx10^(-34)J - s, c = 3xx10^(8)m//s` `E_(ph)=(hc)/(lambda)=(6.63xx10^(-34)xx3xx10^(8))/(10^(-9)xx1.6xx10^(-19))eV = 1243 eV therefore E_(ph) = 1.243 KeV`.

Discussion

31.	A `1 mu A` beam of protons with a cross - sectional area of `0.5 sq.mm` is moving with a velocity of `3 xx 10^(4) ms^(-1)` . Then charge density of beam isA. `6.6 xx 10^(-4) C//m^(3)`B. `6.6 xx 10^(-5) C//m^(3)`C. `6.6 xx 10^(-6) C//m^(3)`D. None of these
Answer» Correct Answer - B For one second , distance = velocity = `3 xx 10^(4) m//sec` and `Q = i xx 1 = 10^(-6) C`. `"Charge density" = (" charge )/(Volume")` ` = (10^(-6))/( 3 xx 10^(4) xx 0.5 xx 10^(-6))` `= 6.6 xx 10^(-5) C//m^(3)`.

Discussion

32.	The momentum of photon of electromagnetic radiation is `3.3xx10^(-29)kgms^-1`. What is the frequency and wavelength of the waves associated with it ? `h=6.6xx10^(-34)Js`.
Answer» Here, `p=3.3xx10^(-29) kg ms^-1, v=? , lambda=?` Energy of a photon `=hv=mc^2=(mv)xxc=pc` or `v=(pc)/h=(3.3xx10^(-29)xx3xx10^8)/(6.6xx10^(-34))=1.5xx10^(13)Hz` `lambda=c/v=(3xx10^8)/(1.5xx10^(13))=2xx10^-5m`

Discussion

33.	Why should gases be insulators at ordanary pressures and start conducting at very low pressures ?
Answer» At ordanary pressure , only very positive ions and electrons are produced are produced by the ionisation of gas moleculews. They are not able to reach the respective electrodes and becaome insulators . At low pressure , density decreases and mean free path becomes large . So at high voltage , they acquire sufficent amount energy and they collide with molcules for further ionnisation . Due to this , the number of ions in a gas increases and it becomes a conductor .

Discussion

34.	The wavelength of electromagnetic radiation is doubled. What will happen to the energy of the photons of the new radiation ?
Answer» ` E = (hc)/lambda rArr E prop 1/lambda `, since wave length of photon is doubled . Its energy becomes halved .

Discussion

35.	Statement-1: The photoelectrons produced by a monochromatic light beam incident on a metal surface have a spread in their kinetic energies. Statement-2: The work function of the metal varies as a function of depth form the surface.A. Statement-1 is true, Statement-2 is true and Statement-2 is correct explanation of Statement-1.B. Statement-1 is true, Statement-2 is true, but Statement-2 is not a correct explanation of Statement-1.C. Statement-1 is true, but Statement-2 is falseD. Statement-1 is false, but Statement-2 is true.
Answer» Correct Answer - a Both statement-1 and statement-2 are true and statement-2 is the correct explanation of the statement-1

Discussion

36.	Every metal has a definite work function . Why do all photoelectrons not come out with the same energy if incident radiation is monochromatic ? Why is there an energy distribution of photoelectrons ?
Answer» Because all the electrons in the metal do not belong to same level but they occupy a continous band of levels , therefore for the given incident radiation , electrons come out come from different levels with different energies . .

Discussion

37.	Assertion : Though light of a single frequency (monochromatic) is incident on a metal , the energies of emitted photoelectrons are different. Reason : The energy of electrons emitted from inside the metal surface is lost in collision with the other atoms in the metal.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If assertion and reason both are false.
Answer» Correct Answer - A When a light of single frequency falls on the electron of inner layer of metal , then this electron comes out of the metal , then this electron comes out of the metal surface a large number of collisions with atom of its upper layer.

Discussion

38.	What is the stopping potential when the metal with work function `0.6 eV` is illuminated with the light of `2 eV` ?A. `2.6 V`B. `3.6 V`C. `0.8 V`D. `1.4 V`
Answer» Correct Answer - D `V_(0) = ((E - W_(0)))/ ( e ) = (( 2eV - 0.6 eV))/( e ) = 1.4 V`

Discussion

39.	When a metallic surface is illuminated with radiation of wavelength `lambda`, the stopping potential is `V`. If the same surface is illuminated with radiation of wavelength `2 lambda` , the stopping potential is `(V)/(4)`. The threshold wavelength surface is :A. ` 3 lambda`B. `4 lambda`C. `5 lambda`D. `(5)/(2) lambda`
Answer» Correct Answer - A Einstein `P.E.` equation Case I : `eV = (hc)/(lambda) - (hc)/(lambda_(0))` ….(i) Case II : `eV//4 = (hc)/( 2lambda) - (hc)/(lambda_(0))` …(ii) From equations (i) and (ii) `rArr 4 = ((1)/(lambda))/((1)/(2 lambda)) - ((1)/(lambda_(0)))/((1)/(lambda))` On solving `lambda_(0) = 31`

Discussion

40.	Light of wavelength `1824 Å`, incident on the surface of a metal , produces photo - electrons with maximum energy `5.3 eV`. When light of wavelength `1216 Å` is used , maximum energy of photoelectrons is `8.7 eV`. The work function of the metal surface isA. `3.5 eV`B. `13.6 eV`C. `6.8 eV`D. `1.5 eV`
Answer» Correct Answer - D `E = W_(0) + K_(max)`. "From the given data" `E is 6.78 eV` ( for `lambda = 1824 Å)` or `10.17 ev ( for lambda = 1216 Å)` `:. W_(0) = E - K_(max) = 6.78 - 5.3 = 1.48 eV` or `W_(0) = 10.17 - 8.7 = 1.47 eV`.

Discussion

41.	The work function of a metallic surface is `5.01 eV`. The photo - electrons are emitted when light of wavelength `2000 Å` falls on it . The potential difference applied to stop the fastest photo - electrons is `[ h = 4.14 xx 10^(-15) eV sec]`A. `1.2 "volts"`B. `2.24 "volts"`C. `3.6 "volts"`D. `4.8 "volts"`
Answer» Correct Answer - A Energy of incident light `E = (12375)/(2000) = 6.18 eV` `"According to relation" E = W_(0) + eV_(0)` `rArr V_(0) = ((E_(0)- W_(0)))/(e ) = ((6.18 eV - 5.01 eV))/( e )` `= 1.17 V ~~ 1.2 V`

Discussion

42.	The work function of metal is `1 eV`. Light of wavelength `3000 Å` is incident on this metal surface . The velocity of emitted photo - electrons will be
Answer» Correct Answer - `v_"max"=1xx10^6` m/s `E=phi+K_"max"`

Discussion

43.	The work function of a metallic surface is `5.01 eV`. The photo - electrons are emitted when light of wavelength `2000 Å` falls on it . The potential difference applied to stop the fastest photo - electrons is `[ h = 4.14 xx 10^(-15) eV sec]`A. 1.2 voltB. 2.24 voltC. 3.6 voltD. 4.8 volt
Answer» Correct Answer - 1

Discussion

44.	Assertion : Photosensitivity of a metal is high if its work function is small. Reason : Work function `= hf_(0)` where `f_(0)` is the threshold frequency.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If both assertion and reason are false.
Answer» Correct Answer - B (b) : Work function is the minimum energy required to eject the photoelectron from photosensitive metal. Hence for metal to be photosensitive, the work function should be small. Work function `=hv_(0)`, where `v_(0)` is the threshold frequency.

Discussion

45.	In photoelectric effect, stopping potential depends onA. frequency of incident lightB. nature of the emitter materialC. intensity of incident lightD. both (a) and (b)
Answer» Correct Answer - D (d) : The stopping potential depends on frequency of incident light and the nature of the emitter material. For a given frequency of incident light, it is independent if its intensity.

Discussion

46.	When a metal surface is exposed with a monochromatic light, then all the photoelectrons are not emitted from the metal surface with the same kintetic energy. Why?
Answer» In photoelectric emission, the maximum KE is aquired by that electron which is most loosely bound to the metal. Since other electrons of metal require different energies for their emission, hence the different photoelectrons are emitted with different energy.

Discussion

47.	When a metallic surface is illuminated with radiation of wavelength `lambda`, the stopping potential is `V`. If the same surface is illuminated with radiation of wavelength `2 lambda` , the stopping potential is `(V)/(4)`. The threshold wavelength surface is :A. `4lambda`B. `5lambda`C. `5lambda//2`D. `3lamda`
Answer» Correct Answer - d Case(i), `eV=(hc)/(lambda)-(hc)/(lambda_(0))..........(i)` Case (ii), `(eV)/4=(hc)/(2lambda)-(hc)/(lambda_(0))` or `eV=(4hc)/(2lambda)-(4hc)/(lambda_(0)).......(ii)` From (i) and (ii), `(hc)/(lambda)-(hc)/(lambda_(0))=(4hc)/(2lambda)-(4hc)/(lambda_(0))` `1/lambda-1/(lambda_(0))=2/lambda-4/(lambda_(0)) or 3/(lambda_(0)) =1/lambda or lambda_(0)=3lambda`

Discussion

48.	The work function of metal is `1 eV`. Light of wavelength `3000 Å` is incident on this metal surface . The velocity of emitted photo - electrons will beA. 10 m/sB. `1xx10^3` m/sC. `1xx10^4` m/sD. `1xx10^6` m/s
Answer» Correct Answer - 4

Discussion

49.	The work function of a metal is `4.2 eV` , its threshold wavelength will beA. 4000 ÅB. 3500 ÅC. 2955 ÅD. 2500 Å
Answer» Correct Answer - 3

Discussion

50.	For given photosenstive material, the photoelectric current is.............on decreasing the wavelength of incident radiation, without any change in intensity of radiation.
Answer» Correct Answer - not affected,

Discussion

Explore topic-wise InterviewSolutions in .

The wavelength of a photon needed to remove a proton from a nucleus which is bound to the nucleus with 1MeV energy is nearlyA. `1.2nm`B. `1.2xx10^(-3)nm`C. `1.2xx10^(-6) nm`D. `.12xx10^(1)nm`

Energy of `K` - shell electron be `-40000 eV`. If `60000 V` potential is applied at Coolidge tube then which of the following `X` - rays will get form ?A. ContinuousB. White `X` - raysC. Continuous and all series of characteristicD. None of these

The wavelength of a photon needed to remove a proton from a nucleus which is bound to the nucleus with 1MeV energy is nearlyA. 1.2 nmB. `1.2xx10^(-3)` nmC. `1.2xx10^(-6)`nmD. `1.2xx10` nm

What determines the hardness of the `X` - rays obtained from the Coolidge filament ?A. Current in the filamentB. Pressure of air in the tubeC. Nature of targetD. Potential difference between cathode and target

It is harder to remove free electron from copper than from sodium. Which has higher work function?

What determines the maximum velocity of the photoelectrons?

A photon of wavelength `6630 Å` is incident on a totally reflecting surface . The momentum delivered by the photon is equal toA. `6.63 xx 10^(-27) kg - m//sec`B. `2 xx 10^(-27) kg - m//sec`C. `10^(-27) kg - m//sec`D. None of these

The work function of copper is 4.0 eV. If two photons, each of energy 2.5 eV strike with some electrons of copper, will the emission be possible?

An electron and a photon have same wavelength `lambda`, what is ratio of their kinetic energies ?A. `h/(m_eclambda)`B. `(2m_e clambda)/h`C. `1:1`D. `h/(2m_eclambda)`

The number of photons `(lambda = 6630 Å)` that strike per second on a totoally reflecting screen (as shown in figure), so that a force of `1N` is exerted on the screen, is approximentely). A. `10^23`B. `10^27`C. `10^25`D. `10^26`

If the cathode - anode potential difference in an `X` - ray tube be `10^(5) V` , then the maximum energy of `X` - ray photon can beA. `10^(5) J`B. `10^(5) MeV`C. `10^(-1) MeV`D. `10^(5) KeV`

If the potential difference between the anode and cathode of the `X` - ray tube is increases A. The peaks at `R` and `S` would move to shorter wavelengthB. The peaks at `R` and `S` would remain at the same wavelengthC. The cut off wavelength at `P` would decreaseD. `(b)` and `( c)` both are correct

Calculate the energy of an electron having de-Broglie wavelength 5500Å. Given, `h=6.62xx10^(-34)Js, m_(e)=9.1xx10^(-31)kg`.

Find the (a) maximum frequency and (b) minimum wave-length of X-rays produced by 30 kV electrons. Given, `h=6.63xx10^(-34)Js`.

The kinetic energy of an electron is `5 eV`. Calculate the de - Broglie wavelength associated with it `( h = 6.6 xx 10^(-34) Js , m_(e) = 9.1 xx 10^(-31) kg)`A. `5.47 Å`B. `109 Å`C. `2.7 Å`D. None of these

Calculate the (a) momentum and (b) de-Broglie wavelength of the electrons accelerated through a potential difference of 56V. Given, `h=6.63xx10^(-34)Js, m_(e)=9.1xx10^(-31)kg, e=1.6xx10^(-19)C`.

What is the (a) momentum (b) speed and (c) de-Broglie wavelength of an electron with kinetic energy of 120 eV. Given `h=6.6xx10^(-34)Js, m_(e)=9xx10^(-31)kg , 1eV=1.6xx10^(-19)J`.

The equivalent wavelength of a moving electron has the same value as that of a photon having an energy of `6xx10^(-17)J`. Calculate the momentum of the electron.

A particle which has zero rest mass and non - zero energy and momentum must travel with a speedA. Equal to c, the speed of light in vacuumB. Tending to infinityC. Greater than cD. Less than c

The momentum of a photon having energy equal to the rest energy of an electron isA. ZeroB. `2.73xx10^(-22) "kg ms"^(-1)`C. `1.99xx10^(-24) "kg ms"^(-1)`D. Infinite

What is the momentum

What is the Speed

Monochromatic light of frequency `6.0 xx 10^(14) Hz` is produced by a laser. The power emitted is `2 xx 10^(-3)` w. The number of photons emitted, on the average, by the sources per second is

Monochromatic light of frequency `6.0xx10^(14)` Hz is produced by a laser. The power emitted is `2.0 xx10^(-3)` W(a) What is the energy of a photon in the light beam ? (b ) How many photons per second , on an average , are emitted by the source ?

What is the de Broglie wavelength of A dust particle of mass `1.0xx10^(-9)` kg drifting with a speed of 2.2 m/s ?

An electron and a photon each has a wavelength of 1.0nm. Find (a) their momenta, (b) the energy of the photon , and © the kinetic energy electron. Take `h=6.63xx10^(-34) Js`.

An electron and a photon each has a wavelength of 1.0nm. Find (a) their momenta, (b) the energy of the photon , and © the kinetic energy electron. Take `h=6.63xx10^(-34) Js`.A.B.C.D.

An electron and a photon each have a wavelength 1.00 nm. Find (i) their momenta, (ii) the energy of the photon and (iii) the kinetic energy of electron.

A `1 mu A` beam of protons with a cross - sectional area of `0.5 sq.mm` is moving with a velocity of `3 xx 10^(4) ms^(-1)` . Then charge density of beam isA. `6.6 xx 10^(-4) C//m^(3)`B. `6.6 xx 10^(-5) C//m^(3)`C. `6.6 xx 10^(-6) C//m^(3)`D. None of these

The momentum of photon of electromagnetic radiation is `3.3xx10^(-29)kgms^-1`. What is the frequency and wavelength of the waves associated with it ? `h=6.6xx10^(-34)Js`.

Why should gases be insulators at ordanary pressures and start conducting at very low pressures ?

The wavelength of electromagnetic radiation is doubled. What will happen to the energy of the photons of the new radiation ?

Every metal has a definite work function . Why do all photoelectrons not come out with the same energy if incident radiation is monochromatic ? Why is there an energy distribution of photoelectrons ?

What is the stopping potential when the metal with work function `0.6 eV` is illuminated with the light of `2 eV` ?A. `2.6 V`B. `3.6 V`C. `0.8 V`D. `1.4 V`

The work function of metal is `1 eV`. Light of wavelength `3000 Å` is incident on this metal surface . The velocity of emitted photo - electrons will be

The work function of a metallic surface is `5.01 eV`. The photo - electrons are emitted when light of wavelength `2000 Å` falls on it . The potential difference applied to stop the fastest photo - electrons is `[ h = 4.14 xx 10^(-15) eV sec]`A. 1.2 voltB. 2.24 voltC. 3.6 voltD. 4.8 volt

In photoelectric effect, stopping potential depends onA. frequency of incident lightB. nature of the emitter materialC. intensity of incident lightD. both (a) and (b)

When a metal surface is exposed with a monochromatic light, then all the photoelectrons are not emitted from the metal surface with the same kintetic energy. Why?

The work function of metal is `1 eV`. Light of wavelength `3000 Å` is incident on this metal surface . The velocity of emitted photo - electrons will beA. 10 m/sB. `1xx10^3` m/sC. `1xx10^4` m/sD. `1xx10^6` m/s

The work function of a metal is `4.2 eV` , its threshold wavelength will beA. 4000 ÅB. 3500 ÅC. 2955 ÅD. 2500 Å

For given photosenstive material, the photoelectric current is.............on decreasing the wavelength of incident radiation, without any change in intensity of radiation.