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 · 25 Jul · Shift 2 · Q65
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 · 25 Jul · Shift 2 · Q65

Dual Nature of Radiation question

2021 · 25 Jul · Shift 2 · Q65

JEE MainPhysicsDual Nature of RadiationNumerical+4 / −1
A light beam of wavelength 500 nm is incident on a metal having work function of 1.25 eV, placed in a magnetic field of intensity B. The electrons emitted perpendicular to the magnetic field B, with maximum kinetic energy are bent into circular are of radius 30 cm. The value of B is ‾×\underline{\hspace{2cm}}\times​× 10 −-− 7 T. Given hc = 20 ×\times× 10 −-− 26 J-m, mass of electron = 9 ×\times× 10 −-− 31 kg
Numerical answer
View written solutionFree

Correct answer: 125

  1. Energy of incident photon

Given:

  • Wavelength, λ=500 nm=5×10−7 m\lambda = 500\ \text{nm} = 5 \times 10^{-7}\ \text{m}λ=500 nm=5×10−7 m
  • hc=20×10−26 J mhc = 20 \times 10^{-26}\ \text{J m}hc=20×10−26 J m

Photon energy:

E=hcλ=20×10−265×10−7=4×10−19 JE = \frac{hc}{\lambda} = \frac{20 \times 10^{-26}}{5 \times 10^{-7}} = 4 \times 10^{-19}\ \text{J}E=λhc​=5×10−720×10−26​=4×10−19 J
  1. Convert work function into joules

Work function:

ϕ=1.25 eV\phi = 1.25\ \text{eV}ϕ=1.25 eV

Using 1 eV=1.6×10−19 J1\ \text{eV} = 1.6 \times 10^{-19}\ \text{J}1 eV=1.6×10−19 J,

ϕ=1.25×1.6×10−19=2×10−19 J\phi = 1.25 \times 1.6 \times 10^{-19} = 2 \times 10^{-19}\ \text{J}ϕ=1.25×1.6×10−19=2×10−19 J
  1. Maximum kinetic energy of emitted electron

By photoelectric equation,

Kmax⁡=E−ϕ=4×10−19−2×10−19=2×10−19 JK_{\max} = E - \phi = 4 \times 10^{-19} - 2 \times 10^{-19} = 2 \times 10^{-19}\ \text{J}Kmax​=E−ϕ=4×10−19−2×10−19=2×10−19 J
  1. Find maximum speed of electron

Using

Kmax⁡=12mv2K_{\max} = \frac{1}{2}mv^2Kmax​=21​mv2

So,

2×10−19=12(9×10−31)v22 \times 10^{-19} = \frac{1}{2}(9 \times 10^{-31})v^22×10−19=21​(9×10−31)v2 v2=4×10−199×10−31=49×1012v^2 = \frac{4 \times 10^{-19}}{9 \times 10^{-31}} = \frac{4}{9} \times 10^{12}v2=9×10−314×10−19​=94​×1012 v=23×106 m/sv = \frac{2}{3} \times 10^6\ \text{m/s}v=32​×106 m/s
  1. Use circular motion in magnetic field

For motion perpendicular to magnetic field,

r=mveBr = \frac{mv}{eB}r=eBmv​

Hence,

B=mverB = \frac{mv}{er}B=ermv​

Given:

  • m=9×10−31 kgm = 9 \times 10^{-31}\ \text{kg}m=9×10−31 kg
  • v=23×106 m/sv = \frac{2}{3} \times 10^6\ \text{m/s}v=32​×106 m/s
  • e=1.6×10−19 Ce = 1.6 \times 10^{-19}\ \text{C}e=1.6×10−19 C
  • r=30 cm=0.3 mr = 30\ \text{cm} = 0.3\ \text{m}r=30 cm=0.3 m

Substitute:

B=9×10−31×23×1061.6×10−19×0.3B = \frac{9 \times 10^{-31} \times \frac{2}{3} \times 10^6}{1.6 \times 10^{-19} \times 0.3}B=1.6×10−19×0.39×10−31×32​×106​ B=6×10−254.8×10−20=1.25×10−5 TB = \frac{6 \times 10^{-25}}{4.8 \times 10^{-20}} = 1.25 \times 10^{-5}\ \text{T}B=4.8×10−206×10−25​=1.25×10−5 T
  1. Match with required format

We need

B=?‾×10−7 TB = \underline{\quad ? \quad} \times 10^{-7}\ \text{T}B=?​×10−7 T

Now,

1.25×10−5=125×10−71.25 \times 10^{-5} = 125 \times 10^{-7}1.25×10−5=125×10−7

So the required integer is:

125\boxed{125}125​
PreviousNext

More from Dual Nature of Radiation

  • In a photoelectric experiment ultraviolet light of wavelength 280 nm is used with lithium cathode having work function ϕ= 2.5 eV. If the wavelength of incident light is switched to 400 nm, find out the change in the stopping…2021 · MCQ
  • The de-Broglie wavelength of a particle having kinetic energy E is λ. How much extra energy must be given to this particle so that the de-Broglie wavelength reduces to 75% of the initial value?2021 · MCQ
  • Given below are two statements : one is labeled as Assertion A and the other is labelled as Reason R. Assertion A : An electron microscope can achieve better resolving power than an optical microscope. Reason R : The de Broglie's…2021 · MCQ
  • The recoil speed of a hydrogen atom after it emits a photon in going from n = 5 state to n = 1 state will be :2021 · MCQ
  • Two stream of photons, possessing energies equal to twice and ten times the work function of metal are incident on the metal surface successively. The value of ratio of maximum velocities of the photoelectrons emitted in the two respective…2021 · Numerical
  • In a photoelectric experiment, increasing the intensity of incident light :2021 · MCQ
  • A monochromatic neon lamp with wavelength of 670.5 nm illuminates a photo-sensitive material which has a stopping voltage of 0.48 V. What will be the stopping voltage if the source light is changed with another source of wavelength of…2021 · MCQ
  • A particle of mass 9.1 × 10 − 31 kg travels in a medium with a speed of 106 m/s and a photon of a radiation of linear momentum 10 − 27 kg m/s travels in vacuum. The wavelength of photon is ​ times the…2021 · Numerical