InterviewSolution
Saved Bookmarks
InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 201. |
If first excitation potential of a hydrogen-like atom is `V` electron volt, then the ionization energy of this atom will be:A. V electron voltB. `(3V)/4` electron voltC. `(4V)/3` electron voltD. cannot be calculated by given information |
|
Answer» Correct Answer - C First excitation energy `=RhC(1/(1^(2))-1/(2^(2))) =RhC 3/4` `:. 3/4 RhC =V e.v` `:. RhC=(4V)/3 e.v.` |
|
| 202. |
The first excitation potential of `He^(+)` ion is `n`, and the ionization potential of `Li^(++)` ion is the `m` then find out value of `(m)/(n)` |
| Answer» Correct Answer - `3` | |
| 203. |
A neutron moving with a speed v strikes a hydrogen atom in ground state moving toward it with the same speed. Find the minimum speed of the neutron for which ineleastic (completely or partially) collision may take place .The mass of neutron = mass of hydrogen `= 1.67 xx 10^(-27)kg` |
|
Answer» Correct Answer - `4` In case of inelastic callision, the `H`-atom will be in excited state. The minimum energy will be when it is perfectly inelastic is, the two `(n)` and `(H)` both come to rest will move will same velocity i.e., `nV+vH " "rarrn+H` from energy conservation `2xx(1)/(2) mv^(2)=E " " v=sqrt((E)/(m))`, `v=sqrt((10.2e)/(1.67xx10^(-27)))=3.13xx10^(4)m//s` where `E` is the excitation energy |
|
| 204. |
An electron is an excited state of `Li^(2 + )`ion has angular momentum `3h//2 pi ` . The de Broglie wavelength of the electron in this state is `p pi a_(0) (where a_(0) ` is the bohr radius ) The value of p is |
|
Answer» Correct Answer - `2` `L=(nh)/(2pi)=(3h)/(2pi)` `n=3` `lambda=(h)/(p)=(h)/(mv)=(h2pir)/(3h)=(2pir)/(3)` `r=a_(0)(n^(2))/(Z)` `lambda=(2pi)/(3)a_(0)(n^(2))/(Z)=(2pi)/(3)a_(0)(3^(2))/(3)=2pia_(0)` |
|
| 205. |
Consider atoms `H, He^(+), Li^(++)` in their ground states. If `L_(1), L_(2)` and `L_(3)` are magnitude of angular momentum of their electrons about the nucleus respectively then:A. `L_(1) = L_(2) = L_(3)`B. `L_(1) gt L_(2) gt L_(3)`C. `L_(1) lt L_(2) lt L_(3)`D. `L_(1) = L_(2) = L_(3)` |
|
Answer» Correct Answer - A Angular momentum `= (nh)/(2pi)` i.e., same for all. |
|
| 206. |
Consider atoms `H, He^(+), Li^(++)` in their ground states. If `L_(1), L_(2)` and `L_(3)` are magnitude of angular momentum of their electrons about the nucleus respectively then:A. `L_(1)=L_(2)=L_(3)`B. `L_(1) gt L_(2) gt L_(3)`C. `L_(1) lt L_(2) lt L_(3)`D. `L_(1)=L_(2)=L_(3)` |
|
Answer» Correct Answer - A Angular momentum `=(nh)/(2pi)rArr ` i.e. same for all |
|
| 207. |
Assertion: `alpha` and `beta` particles are accelerated through same potential difference. Finally both particles have sma linear momentum. Reason: Linear momentum `= sqrt(KE xx 2 xx mass)`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 is false but reason is true. |
|
Answer» Correct Answer - D `p = sqrt(2mKE)` `palpha = sqrt(2m_(alpha)KE)` `pbeta = sqrt(2m_(beta) KE)` `kE` is for `alpha` prticles is two times the `KE` of `beta` particles. But `m_(alpha) gt gt gt m_(beta)` `rArr p_(alpha) != p_(beta)` |
|
| 208. |
Which one of the series of hydrogen spectrum is in the visible region ?A. Lyman seriesB. Balmer seriesC. Paschen seriesD. Bracket series |
|
Answer» Correct Answer - B Balman series lies in the visible region. |
|
| 209. |
The visible region of hydrogen spectrum was first studied byA. LymanB. BalmerC. PfundD. Brackett |
| Answer» Correct Answer - B | |
| 210. |
The Fine structure of hydrogen spectrum can be explained byA. the presence of neutrons in the nucleusB. the finite size of nucleusC. the obtical angular momentum of electronsD. the spin angular momentum of the electron |
| Answer» Correct Answer - D | |
| 211. |
Which of the following parameters are the same for all hydrogen like atoms and ions in their ground state?A. radius of the orbitB. speed of the electronC. energy of the atomD. orbital angular momentum of electrons |
| Answer» Correct Answer - D | |
| 212. |
A gas of hydrogen - like ion is perpendicular in such a way that ions are only in the ground state and the first excite state. A monochromatic light of wavelength `1216 Å` is absorved by the ions. The ions are lifted to higher excited state and emit emit radiation of six wavelength , some higher and some lower than the incident wavelength. Find the principal quantum number of the excited state identify the nuclear charge on the ions . Calculate the values of the maximum and minimum wavelengths.A. `51 eV`B. `54.4 eV`C. `63.75 eV`D. `69 eV`. |
| Answer» Correct Answer - C | |
| 213. |
An electron, in a hydrogen like atom , is in excited state. It has a total energy of -3.4 eV, find the de-Broglie wavelength of the electron. |
|
Answer» a. energy of electron in hydrogen-like atom, `E_(n) = - (Z^(2) Rhc)/(n^(2)) = - 3.4 e V` Kinetic energy of electron in hydrogen-like atom is equal to negative of total energy i.e. , ` E_(k) = - E_(n) = - (-3.4 e V) = + 3.4 e V` (ii) The de Broglie wavelength of electron . `lambda = (h)/(p) = (h)/sqrt(2 mE_(k))` `= (6.63 xx 10^(-34))/sqrt{{2 xx 9.1 xx 10^(-31) xx 3.4 xx 1.6 xx 10^(-19)}} `= 6.66 Å` |
|
| 214. |
The shorted wavelength of X- rays emitted from an X- rays tube depends onA. the current in the tubeB. the voltage applied to the tubeC. the nature of the gas in the tubeD. the atomic number of the target meterial |
|
Answer» Correct Answer - A Shortest wavelength or cut-off wavelength depends only open the voltage applied in the Coolidge tube. |
|
| 215. |
If element with particle quantum number `ngt4` were not allowed in nature , the number of possible element would beA. 60B. 32C. 4D. 64 |
|
Answer» Correct Answer - D The maximum number of electron in an orbit is `2 n^(2)`. Since `n gt4` is not possible , therefore the maximum number of electron that can be in the first four orbit are `2 (1)^(2) + 2 (2)^(2) + 2 (3)^(2) + 2 (4) = 2 + 8 + 18 + 32 = 60` Therefore, possible element are`60` |
|
| 216. |
The orbital velocity of electron in the ground state is `v`. If the electron is excited to enegry staet `- 0.54 eV` its orbital velocity will beA. `v`B. `(v)/(3)`C. `(v)/(5)`D. `(v)/(7)` |
|
Answer» Correct Answer - C `E_(n) = - (13.6)/(n^(2)) implies n^(2) = - (13.6)/(-0.54)` or `n^(2) = 25.2 or n = 5` (nearly) `As v prop 1//n , so v_(n) = (v)/(5)` |
|
| 217. |
Figure shows the variation of frequency of a characterstic x-ray and atomic number. (i) Name the characterstics x-ray (ii) Find the energy of photon emitted when this x-ray is emitted by a metal having `z=101` |
|
Answer» Correct Answer - (i) `K_(alpha)` (ii) `102 KeV` From graph `gammaK(Z-1)^(2)` where `K` is constant `sqrt(v)=a(z-1) a=sqrt(K)` So its `K_(alpha)-x-ray` Energy of photon `=13.6xx(z-b)^(2)((1)/(n_(1)^(2))-(1)/(n_(2)^(2)))` For `K_(alpha)n_(1)=1` `n=2` `E=13.6xx(100^(2))((1)/(n_(1)^(2))-(1)/(n_(2)^(2)))` `=102 KeV` |
|
| 218. |
Find the wavelength of the `K_(alpha)` line in copper `(Z=29)`, if the wave length of the `K_(alpha)` line in iron `(Z=26)` is known to be equal to `193 "pm"` (Take `b=1`) |
| Answer» Correct Answer - `lambda_(1)=((26-1)/(29-1))^(2)193"pm"=154"pm"` | |
| 219. |
A hydrogen like atom (atomic number Z) is in a higher excited state of quantum number n. The excited atom can make a transition to the first excited state by successively emitting two photons of energy 10.2 eV and 17.0 eV, respectively. Alternatively, the atom from the same excited state can make a transition to the second excited state by successively emitting two photons of energies 4.25 eV and 5.95 eV, respectively Determine the values of n and Z. (lonization energy of H-atom = 13.6 eV) |
|
Answer» Correct Answer - `n=6, Z=3` From the given conditions : `E_(n)-E_(2)=(10.2+17)eV=27.2 eV` and `E_(n)-E_(3)=(4.25+5.95)eV=10.2 eV` Equations (i)-(ii) gives. `E_(3)-E_(2)=17.0 eV` or `Z^(2)(13.6)((1)/(4)-(1)/(9))=17.0` `rArrZ^(2)(13.6)(5//36)=17.0` `rArr Z^(2)=9` or `Z=3` From equation (i), `Z^(2)(13.6)((1)/(4)-(1)/(9))=27.2` or `(3)^(2)(13.6)((1)/(4)-(1)/(n^(2)))=27.2` or `((1)/(4)-(1)/(n^(2)))=0.222` or `(1)/(n^(2))=0.0278` or `n^(2)=36` `:. n=6` |
|
| 220. |
For the first member of Balmer series of hydrogen spectrum, the wavelength is `lamda`. What is the wavelength of the second member?A. `(20lambda)/(27)`B. `(3lambda)/(16)`C. `(5lambda)/(36)`D. `(3lambda)/(4)` |
|
Answer» Correct Answer - A `(1)/(lambda_(1))=R ((1)/(4)-(1)/(9)) rArrlambda_(1)=(4xx9)/(5R)` Similarly `(1)/(lambda_(2))=R((1)/(4)-(1)/(4^(2))) rArrlambda_(2)=(16)/(3R)=(16)/(3)xx(5lambda)/(4xx9)=(20)/(27)lambda` |
|
| 221. |
If the wavelength of the first member of Balmer series of hydrogen spectrum is `6564A^(@)` , the wavelength of second member of Balmer series will be:A. `1215A^(@)`B. `4848A^(@)`C. `6050A^(@)`D. data given is insufficient to calculate the value |
|
Answer» Correct Answer - B `1/lambda=RZ^(2)(1/n_(1)^(2)(-1)/n_(2)^(2)) "for Balmer sereis"` `n_(1)=2` and `n_(2)=3,4,5....` |
|
| 222. |
The seond line of Balmer series has wavelength `4861 Å` The wavelength o fthe first line Balmer series isA. `1216 Å`B. `6563 Å`C. `4340 Å`D. `4101 Å` |
|
Answer» Correct Answer - B For first member, `(1)/(lambda_(1)) = R[(1)/(2^(2))-(1)/(3^(2))]` or `(1)/(lambda_(1)) = R[(1)/(4)-(1)/(9)]` or `(1)/(lambda) = R [(9-4)/(36)]` or `lambda_(1) = (36)/(5R)` For second mumber, `(1)/(lambda) = R[(1)/(2^(2))-(1)/(4^(2))]` or `(1)/(lambda_(2)) = R[(1)/(4)-(1)/(16)]` or `(1)/(lambda_(2)) = R[(4-1)/(16)]` or `(1)/(lambda_(2)) = (3R)/(16)` or `lambda_(2) = (16)/(3R)` Now, `(lambda_(1))/(lambda_(2)) = (36)/(5R) xx (3R)/(16)` or `(lambda_(1))/(lambda_(2)) = (27)/(20)` or `lambda_(1) = (27)/(20) xx lambda_(2) = (27)/(20)xx 4861 Å = 6562.4 Å` . |
|
| 223. |
An X-ray tube is operated at `50 kV and 20 m A`. The target material of the tube has mass of `1 kg` and specific heat `495 J kg ^(-1) `^(@)C^(-1)`. One perent of applied electric power is converted into X-rays and the remaining energy goes into heating the target. Then,A. the average rate of rise of temperature of the target would be `2_(0)C//sec`B. the minimum wavelength of the `X`-rays emitted is about `0.25xx10^(-10)m`C. a suitable target material must have a high melting temperatureD. a suitable target material must have thermal conductivity |
|
Answer» Correct Answer - B Power reaching the target `p=Vi=1000W` . The rate of heat production =`(99)/(100)xx1000=990W` `P=(dttheta)/(dt)=(mstheta)/(dt)=990Rightarrow(dttheta)/(dt)=2^(0)c//_(s)` and `lambda_(min)=(hc)/(eV)` |
|
| 224. |
`O^(++), C^(+), He^(++)` and `H^(+)` inos are projected on the photographic plate with same velocity in a sepctrograph. Which one will strike farthest ?A. `O^(++)`B. `C^(+)`C. `He^(++)`D. `H_(2)^(+)` |
|
Answer» Correct Answer - B `2r = (2mv)/(qB) rArr 2r prop (m)/(q) rArr (m)/(q)` is maximum for `C^(+)` |
|
| 225. |
On decreasing principal quantum number n, the value of r and v willA. decreaseB. increaseC. r will increase but v will decrease.D. r will decrease but v will increase |
| Answer» Correct Answer - D | |
| 226. |
Assertion: The force of repulsion between atomic nucleus and `alpha`-particle varies with distance according to inverse square law. Reason: Rutherford did `alpha`-particles scattering experiment.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 is false but reason is true. |
|
Answer» Correct Answer - B Rutherford confirmed the repulsive force on `alpha`-particle due to nucleus varies with distance according to inverse square law and that the positive changes are concentraed at the cnetre and not distributed throughout the atom. |
|
| 227. |
As the orbit number increase , the distance between two consecutive orbits in an atom or ion having single electron:A. increasesB. decreasesC. remain the sameD. first increases and then become constant |
| Answer» Correct Answer - A | |
| 228. |
An atomic nucleus containsA. only electronsB. only protonsC. only neutronsD. both proton and neutrons |
| Answer» Correct Answer - D | |
| 229. |
The ratio of the largest to shortest wavelength in Lyman series of hydrogen spectra isA. `(25)/(9)`B. `(17)/(6)`C. `(9)/(5)`D. `(4)/(3)` |
|
Answer» Correct Answer - D For Lyman series `(1)/(lambda_(max)) = R [(1)/(1^(2)) - (1)/(2^(2))] = (3)/(4)R` and `(1)/(lambda_(min)) = R[(1)/(1^(2)) - (1)/(oo^(2))] = (R )/(1) rArr (lambda_(max))/(lambda_(min)) = (4)/(3)` |
|
| 230. |
The ratio of the longest to shortest wavelength in Brackett series of hydrogen spectra isA. `(25)/(9)`B. `(17)/(6)`C. `(9)/(5)`D. `(4)/(3)` |
|
Answer» Correct Answer - A For Bracket series `(1)/(lambda_(max)) = R[(1)/(4^(2)) - (1)/(5^(2))] = (9)/(25 xx 16) R` and `(1)/(lambda_(min)) = R[(1)/(4^(2)) - (1)/(oo^(2))] = (R )/(16) rArr (lambda_(max))/(lambda_(min)) = (25)/(9)` |
|
| 231. |
`1.8 g` of hydrogen is excite by irradiation. The study of spectra indicated that `27 %` of the atoms are in the first excite state, `15 %` of the aton in the second excited, and the rest in the ground state. The ground stat, ionization energy of hydrogen atom is `21.4 xx 10^(-12)`ergs. The total amount of energy that would be evolved when all the atoms neutron to the ground state isA. `782 k J`B. `978 k J`C. `19.63 xx 10^(11) erg`D. `97.87 xx 10^(11) erg` |
|
Answer» Correct Answer - A Amount of energy that would be evolved `= U_(i) - U_(f)` `U_(i) = N_(1) E_(1) + N_(2) E_(2)` `= - 1.61 xx 10^(23) xx (21 xx 10^(-12))/(3^(2)) - (2.9 xx 10^(23) xx 21.7 xx 10^(-12))/2^(2)` `= - 3.88 xx 10^(11) - 15.75 xx 10^(11)` `= - 19.63 xx 10^(11)` erg `U_(f) = (N_(1) + N_(2)) E_(0)` `= - (1.61 xx 10^(23) + 2.9 xx 10^(23)) 21.7 xx 10^(-12)` `= - 97.87 xx 10^(11)`erg Energy evolved `U_(i) - U_(f) = (- 19.63 xx 10^(11) + 97.87 xx 10^(11))` `= 78.237 xx 10^(11) = 782 k J` |
|
| 232. |
According to classical theory, the circular path of an electron in Rutherford atom isA. straight lineB. spiralC. circularD. parabolic |
| Answer» Correct Answer - B | |
| 233. |
The ratio of minimum to maximum wavelength in Balmer series isA. `5:9`B. `5:36`C. `1:4`D. `3:4` |
|
Answer» Correct Answer - A `(1)/(lambda) prop ((1)/(n_(1)^(2))-(1)/(n_(2)^(2)))` `(lambda_(min))/(lambda_(max)) = (((1)/(2^(2)) - (1)/(3^(2))))/(((1)/(2^(2)) - (1)/(oo))) = (5)/(9)`a |
|
| 234. |
In an experiment on photoelctric effect, light of wavelength `800 nm` ( less than threshold wavelength) is incident on a cessium plate at the rate of `5.0W`. The potential of the collector plate is made sufficiently positive with respect to the emitter so that the current reaches its saturation value. Assuming that on the average one of every `10^(6)` photons is able to eject a photoelectron, find photo current in the circuit. |
|
Answer» Correct Answer - `(Plambda)/(hcxx10^(6))e A=3.2 mu A` No. of photons `(P)/(E_(lambda))=(P_(lambda))/(hc).s^(-1)` no of photo electron `s//s=(Plambda)/(hc).(1)/(10^(6))` `:.` photo current `=(Plambda)/(hcxx10^(6)).e=(5xx800xx10^(-9)xx(1.6xx10^(-19)))/(6.63xx10^(-34)xx3xx10^(8)xx10^(6))A=3.2 mu A.` |
|
| 235. |
When an electron moving at a high speed strikes a metal surface, which of the following are possible? (i) The entire energy of the electron may be converted into an X-ray photon (ii) Any fraction of energy of the electron may be converted into an X-ray photon (iii) The entire energy of the electron may get converted to heat (iv) The electron may undergo elastic collision with the metal surfaceA. the entire energy of the electron may be converted into an X-ray photonB. any fraction of the energy of the electron may be converted into an X-ray photonC. the entire energy of the electron may get converted to heatD. the electron may under go elastic collision with the metal surface. |
|
Answer» Correct Answer - A::B::C::D the electron will not undergo elastic collision with the metal surface |
|
| 236. |
If the voltage across the filament is increased, the cutoff wavelengthA. will increaseB. will decreaseC. will remain unchangedD. will change |
|
Answer» Correct Answer - C The cut off wavelength depends on the accelerating potential difference which is unchanged. Hence, the wavelength will remain unchanged. |
|
| 237. |
The wavelength of a photon and the deBroglie wavelength of an electron and uranium atom are identical. Which one of them will have highest kinetic energyA. PhotonB. electronC. U-atomD. nothing can be predicted |
|
Answer» Correct Answer - A `lambda_(D)=h/p =h/(sqrt(2mE)) rArr E=(h^(2))/(2mlambda^(2)) E prop 1/m` (for electron and uranium) for photon and electron `lambda_("photon") =lambda_("electron")` (given) `p_("photon")=p_("electron")` `(E_("electron"))/(E_("photon"))=(1/2 mv^(2))/(P_("photon")xxC) =((P_("electron")V)/2)/(P_("photon")xxC)=v/(2C)` |
|
| 238. |
The average lifetime for the `n = 3` excited state of a hydrogen-like atom is `4.8 xx 10^(-8) s` and that for the `n = 2 state is 12.8 xx 10^(-8) s`. The ratio of average number of revolution made in the `n = 3` sate before any transition can take place from these state is. |
|
Answer» Correct Answer - 9 Numbersof revolution before transition = `"frequency"` xx times Also, ` "frequency" prop (1)/(n^(3))` `"So required ratio" = ((1//2) 12.8 xx 10^(-8))/((1//3) 4.8 xx 10^(-8)) = 9` |
|
| 239. |
Heat at the rate of 200 W is produced in an X-ray tube operating at 20 kV. Find the current in the circuit. Assume that only a small fraction of the kinetic energy of electron is converted into X-rays. |
|
Answer» Correct Answer - 1 Heat produced //sec`= 200W` `implies I = (200)/(V) = (200)/(20 xx 10^(3)) = 10 mA` |
|
| 240. |
`K_alpha` wavelength emitted by an atom of atomic number Z=11 is `lambda`. Find the atomic number for an atom that emits `K_alpha` radiation with wavelength `4lambda`. (a) Z=6 (b) Z=4 (c ) Z=11 (d) Z=44. |
|
Answer» `lambda_(1)/lambda_(2)=(Z_(2)-1)^(2)/(Z_(1)-1)^(2)(since,1/lambdaalpha(Z-1)^(2))` ` 1/4=(Z_(2)-1)^(2)/(11-1)^(2)`on solving `Z_(2)=6` |
|
| 241. |
A free atom of iron emits `K_alpha` X-rays of energy 6.4 keV. Calculate the recoil kinetic energy of the atom. Mass of and iron atom `= 9.3 xx 10 ^(-26) kg`. |
| Answer» Correct Answer - `K_("recoil")=((6.4xx10^(3)e)/(c ))xx(1)/(2xx(9.3xx10^(26)))J` | |
| 242. |
The potential energy of a particle of mass `m` is given by `V(x) = E_0` when `x= lex le 1 `and `xgt 1` repectively. `lambda_(1) and lambda_(2) ` are the de - Broglie wavelength of the particle, ,if the total energy of particle is `2 E_(0)` find `lambda_(1) // lambda_(2)`A. `sqrt2`B. `1/sqrt2`C. `2:1`D. `4:1` |
|
Answer» Correct Answer - A `lambda_(1)/lambda_(2)=(h)/(p_(1)/(h))=p_(2)/p_(1)=sqrt(2mk_(2))/(sqrt(2mk_(1)))=sqrt(2E_(0))/sqrt(2E_(0)-E_(0))=sqrt2` |
|
| 243. |
If the wavelength of the `n^(th)` line of Lyman series is equal to the de-broglie wavelength of electron in initial orbit of a hydrogen like element `(Z=11)`. Find the value of `n`. |
|
Answer» Correct Answer - `n=24` If the wavelength of the `n^(th)` line of Lyman series is equal to the de-broglie wavelength of electron in initial orbit `(1)/(lambda)=RZ^(2)(1-(1)/((n+1)^(2)))=(1)/((h//mv))=(mv)/(h)=(mvr)/(hr)=` `((n+1)h)/(2pihr)=(n+1)/(2pir)` putting, `r=(r_(0)(n+1)^(2))/(Z)` `(1)/(lambda)=RZ^(2)(1-(1)/((n+1)^(2)))=((n+1)Z)/(2pir_(0)(n+1)^(2))" " rArr(n+1)-(1)/((n+1))=(1)/(2pir_(0)ZR)=(1)/(2pixx11xx0.53xx10^(-10)xx1.1xx10^(7))` `=(1000)/(2pixx12.1xx0.53)=24.8" "rArr(n+1)=25" "rArrn=24` |
|
| 244. |
The potential energy of a particle of mass `m` is given by `V(x) = E_0` when `x= lex le 1 `and `xgt 1` repectively. `lambda_(1) and lambda_(2) ` are the de - Broglie wavelength of the particle, ,if the total energy of particle is `2 E_(0)` find `lambda_(1) // lambda_(2)` |
|
Answer» Correct Answer - `sqrt(2)` For `0 lt x le1, " "KE=2E_(0)-E_(0)=E_(0)` for `x gt 1," " KE=2E_(0)" "(lambda_(1))/(lambda_(2))=(h//P_(1))/(h//P_(2))=(P_(2))/(P_(1))=sqrt((KE_(2))/(KE_(1)))=sqrt(2)` |
|
| 245. |
A photon of `10.2 eV`. Energy collides with a hydrogen atom in ground state inelastically. After few microseconds one more photon of energy `15 e V` collides with the same hydrogen atom. Then what can be detected by a suitable detector.A. one photon of `10.2 eV` and an electron of energy `1.4 eV`B. `2` photon of energy `10.2 eV`C. `2` photons of energy `3.4 eV`D. `1` photon of `3.4 eV` and one electron of `.4 eV` |
|
Answer» Correct Answer - A First photon will excite the atom to `I` excited state, which when returining to ground state will emit a photon of energy `10.2 eV` second photon will ionize the atom (`13.6 eV` will be used up in this process). The extra energy `(=15013.6=1.4 eV)` will be carried by electron as its kinetic energy. So a photon of energy `13.6 eV` and an electron of energy `1.4 eV` will be emitted. |
|
| 246. |
A particle of charge `q_(0)` and of mass `m_(0)` is projected along the `y`-axis at `t=0` from origin with a velocity `V_(0)`. If a uniform electric field `E_(0)` also exist along the `x`-axis, then the time at which debroglie wavelength of the particle becomes half of the initial value is:A. `(m_(0)v_(0))/(q_(0)E_(0))`B. `2(m_(0)v_(0))/(q_(0)E_(0))`C. `sqrt(3) (m_(0)v_(0))/(q_(0)E_(0))`D. `3(m_(0)v_(0))/(q_(0)E_(0))` |
|
Answer» Correct Answer - C Initial de broglie wavelength `=h/(m_(0)v_(0))` after any time t, `lambda=h/(sqrt((m_(0)v_(0))^(2)+(q_(0)E_(0)t)^(2)))` When `lambda` becomes half of the initial value `h/(2m_(0)v_(0))=h/(sqrt(m_(0)v_(0))^(2)+(q_(0)E_(0)t)^(2))=sqrt(3) m_(0)v_(0)=q_(0)E_(0)t rArr t=sqrt(3) (m_(0)v_(0))/(q_(0)E_(0))` |
|
| 247. |
Electron are emitted from an electron gun at almost zero velocity and are accelerated by an electric field `E` through a distance of `1.0m` The electron are now scattered by an atomic hydrogen sample in ground state what should be the minimum value of `E` so that red light of wavelength `656.5 nm` may be emitted by the hydrogen? |
|
Answer» Correct Answer - `[12.1 "volts"//m]` |
|
| 248. |
When an electron acceleration by potential different `U` is bombarded on a specific metal, the emitted X-ray spectrum obtained is shown in figure . If the potential different is reduced to `U//3` , the correct spectrum is B. C. D. |
|
Answer» Correct Answer - B `lambda_(m)` will increase to `3 lambda_(m)` due to derease in the energy of bombarding electrons. Hence , no characteristic X- ray willl be visible , only continuous X- ray willl be produced. |
|
| 249. |
An electron beam accelerated from rest through a potential difference of `5000 V`in vacuum is allowed to impinge on a surface normaly. The incident current is `muA` and if the electrons come to rest on striking the surface the force on it is:A. `1.1924 xx 10^(-8) N`B. `2.1 xx 10^(-8) N`C. `1.6 xx 10^(-8)N`D. `1.6 xx 10^(-6) N` |
|
Answer» Correct Answer - A Energy `=(1)/(2)mv^(2) = 5000eV` `=5000 xx 1.6 xx 10^(-19)` joule `mv = sqrt(2xx5000 xx (16 xx 10^(-19)m))` `= 4 xx 10^(-8) xx sqrt(m)` Number of electron striking per second `=n = (q)/(e) = (It)/(e) = (50xx10^(-6)xx1)/(1.6 xx 10^(-19))` `= 31.25 xx 10^(13)` Force `=` change of momentum per second `=n(mv) = 31.25 xx 10^(13) xx 4xx10^(-8) sqrt(m)` `= 125 xx 10^(5) sqrt(9.1 xx 10^(-31))` `= 1.1924 xx 10^(-8)` newton |
|
| 250. |
The Rydberg constant `R` for hydrogen isA. `R = -((1)/(4pi epsilon_(0))).(2pi^(2)me^(2))/(ch^(2))`B. `R = ((1)/(4pi epsilon_(0))).(2pi^(2)me^(4))/(ch^(2))`C. `R =((1)/(4pi epsilon_(0)))^(2).(2pi^(2)me^(4))/(c^(2)h^(2))`D. `R = ((1)/(4pi epsilon_(0)))^(2).(2pi^(2)me^(4))/(ch^(3))` |
|
Answer» Correct Answer - D `R = (2pi^(2)k^(2)e^(4)m)/(ch^(3)) = ((1)/(4pi epsilon_(0)))^(2) (2pi^(2) me^(4))/(ch^(3))` |
|