Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Dual Nature of Radiation question

2021 · 20 Jul · Shift 1 · Q57
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Physics
  4. /Dual Nature of Radiation
  5. /2021 · 20 Jul · Shift 1 · Q57

Dual Nature of Radiation question

2021 · 20 Jul · Shift 1 · Q57

JEE MainPhysicsDual Nature of RadiationMCQ+4 / −1
The radiation corresponding to 3 →\to→ 2 transition of a hydrogen atom falls on a gold surface to generate photoelectrons. The electrons are passed through a magnetic field of 5 ×\times× 10 −-− 4 T. Assume that the radius of the largest circular path followed by these electrons is 7 mm, the work function of the metal is : (Mass of electron = 9.1 ×\times× 10 −-− 31 kg)
  1. A
    1.36 eV
  2. B
    1.88 eV
  3. C
    0.82 eV
  4. D
    0.16 eV
View written solutionFree

Correct answer: C

  1. Energy of the incident photon

For the hydrogen atom, the energy levels are En=−13.6n2 eVE_n = -\frac{13.6}{n^2}\text{ eV}En​=−n213.6​ eV

For the transition 3→23 \to 23→2, ΔE=13.6(122−132)\Delta E = 13.6\left(\frac{1}{2^2} - \frac{1}{3^2}\right)ΔE=13.6(221​−321​) =13.6(14−19)= 13.6\left(\frac{1}{4} - \frac{1}{9}\right)=13.6(41​−91​) =13.6(536)= 13.6\left(\frac{5}{36}\right)=13.6(365​) =1.89 eV (approximately)= 1.89\text{ eV (approximately)}=1.89 eV (approximately)

So, the incident photon energy is hν=1.89 eVh\nu = 1.89\text{ eV}hν=1.89 eV

  1. Maximum kinetic energy of emitted photoelectrons

The largest radius corresponds to the maximum speed of photoelectrons.

For motion in a magnetic field, r=mveBr = \frac{mv}{eB}r=eBmv​

Hence, v=eBrmv = \frac{eBr}{m}v=meBr​

Given:

  • e=1.6×10−19 Ce = 1.6 \times 10^{-19}\,\text{C}e=1.6×10−19C
  • B=5×10−4 TB = 5 \times 10^{-4}\,\text{T}B=5×10−4T
  • r=7 mm=7×10−3 mr = 7\,\text{mm} = 7 \times 10^{-3}\,\text{m}r=7mm=7×10−3m
  • m=9.1×10−31 kgm = 9.1 \times 10^{-31}\,\text{kg}m=9.1×10−31kg

So, v=(1.6×10−19)(5×10−4)(7×10−3)9.1×10−31v = \frac{(1.6\times10^{-19})(5\times10^{-4})(7\times10^{-3})}{9.1\times10^{-31}}v=9.1×10−31(1.6×10−19)(5×10−4)(7×10−3)​

First calculate numerator: 1.6×5×7=561.6\times5\times7 = 561.6×5×7=56 10−19×10−4×10−3=10−2610^{-19} \times 10^{-4} \times 10^{-3} = 10^{-26}10−19×10−4×10−3=10−26 So, v=56×10−269.1×10−31v = \frac{56\times10^{-26}}{9.1\times10^{-31}}v=9.1×10−3156×10−26​ =569.1×105= \frac{56}{9.1}\times10^5=9.156​×105 ≈6.15×105 m/s\approx 6.15\times10^5\,\text{m/s}≈6.15×105m/s

Now, Kmax⁡=12mv2K_{\max} = \frac{1}{2}mv^2Kmax​=21​mv2 =12(9.1×10−31)(6.15×105)2= \frac{1}{2}(9.1\times10^{-31})(6.15\times10^5)^2=21​(9.1×10−31)(6.15×105)2

v2≈3.78×1011v^2 \approx 3.78\times10^{11}v2≈3.78×1011

Thus, Kmax⁡=0.5×9.1×10−31×3.78×1011K_{\max} = 0.5\times 9.1\times10^{-31}\times 3.78\times10^{11}Kmax​=0.5×9.1×10−31×3.78×1011 ≈1.72×10−19 J\approx 1.72\times10^{-19}\,\text{J}≈1.72×10−19J

Convert to eV: Kmax⁡=1.72×10−191.6×10−19 eVK_{\max} = \frac{1.72\times10^{-19}}{1.6\times10^{-19}}\text{ eV}Kmax​=1.6×10−191.72×10−19​ eV ≈1.07 eV\approx 1.07\text{ eV}≈1.07 eV

  1. Apply photoelectric equation

Using Einstein's photoelectric equation, hν=ϕ+Kmax⁡h\nu = \phi + K_{\max}hν=ϕ+Kmax​

So, ϕ=hν−Kmax⁡\phi = h\nu - K_{\max}ϕ=hν−Kmax​ =1.89−1.07= 1.89 - 1.07=1.89−1.07 =0.82 eV= 0.82\text{ eV}=0.82 eV

  1. Option check
  • A: 1.36 eV1.36\,\text{eV}1.36eV ❌
  • B: 1.88 eV1.88\,\text{eV}1.88eV ❌
  • C: 0.82 eV0.82\,\text{eV}0.82eV ✅
  • D: 0.16 eV0.16\,\text{eV}0.16eV ❌

Therefore, the correct answer is Option C.

PreviousNext

More from Dual Nature of Radiation

  • An electron having de-Broglie wavelength λ is incident on a target in a X-ray tube. Cut-off wavelength of emitted X-ray is :2021 · MCQ
  • A certain metallic surface is illuminated by monochromatic radiation of wavelength λ. The stopping potential for photoelectric current for this radiation is 3V0. If the same surface is illuminated with a radiation of wavelength 2 λ…2021 · Numerical
  • An electron of mass me and a proton of mass mp are accelerated through the same potential difference. The ratio of the de-Broglie wavelength associated with the electron to that with the proton is2021 · MCQ
  • Given below are two statements : Statement I : Two photons having equal linear momenta have equal wavelengths. Statement II : If the wavelength of photon is decreased, then the momentum and energy of a photon will also decrease. In the…2021 · MCQ
  • An X-ray tube is operated at 1.24 million volt. The shortest wavelength of the produced photon will be :2021 · MCQ
  • The de-Broglie wavelength of a proton and α-particle are equal. The ratio of their velocities is :2021 · MCQ
  • An α particle and a proton are accelerated from rest by a potential difference of 200V. After this, their de Broglie wavelengths are λα and λ p respectively. The ratio λα​λp​​…2021 · MCQ
  • An electron of mass me and a proton of mass mp = 1836 me are moving with the same speed. The ratio of their de Broglie wavelength λprotonλelectron​ will be :2021 · MCQ