Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Light of wavelength ejects photo-electrons from a plate of metal whose work-function

Light of wavelength 180 \mathrm{~nm} ejects photo-electrons from a plate of metal whose work-function is 2 \mathrm{eV}. If a uniform magnetic field of 5 \times 10^{-5} \mathrm{Tesla} be applied paralled to the plate, what would be the radius of the path followed by electrons ejected normally from the plates with maximum energy \left(\mathrm{h}=6.62 \times 10^{-34}\right. \text { Joule- }\text { sec, } \mathrm{m}=9.1 \times 10^{-31} \mathrm{~kg} \text { and } \mathrm{e}=1.6 \times 10^{-19} \text { coulomb). }

Option: 1

0.12 \mathrm{~m}


Option: 2

0.08 \mathrm{~m}


Option: 3

0.21 \mathrm{~m}


Option: 4

0.15 \mathrm{~m}


Answers (1)

best_answer

Given that \mathrm{W}=2 \, \mathrm{eV} and  \lambda=180 \times 10^{-9} \mathrm{~m} . 

According to Einstein’s photoelectric equation, maximum kinetic energy of the emitted photo electrons

\mathrm{E}_{\mathrm{k}}^{\max }=\left(\frac{1}{2}\right) \mathrm{mv}_{\max }^2=\frac{\mathrm{hc}}{\lambda}-\mathrm{W}

=\frac{\left(6.62 \times 10^{-34} \mathrm{~J}-\mathrm{S}\right) \times\left(3 \times 10^8 \mathrm{mS}^{-1}\right)}{\left(180 \times 10^{-9} \mathrm{~m}\right)}\approx \left(2 \times 1.6 \times 10^{-19} \mathrm{~J}\right)

=7.8 \times 10^{-19} \text { Joule }

\therefore V_{\max }=\sqrt{\frac{2 E_K^{\max }}{\mathrm{m}}}=\sqrt{\frac{2 \times 7.8 \times 10^{-19}}{9.1 \times 10^{-31}}}=1.3 \times 10^6 \mathrm{~m} / \mathrm{S}

As a magnetic field is applied parallel to the plate and electrons are ejected normally from the plate with maximum velocity \mathrm{V}_{\max }, hence Lorentz force acting on the electrons =\mathrm{Bev}_{\max }. Because direction of Lorentz force will be perpendicular to the direction of motion of electrons, hence electrons will move on a circular path. The radius of circular path is given by

\mathrm{Bev}_{\max }=\frac{\mathrm{mv}_{\max }^2}{r} \text { or } r=\frac{\mathrm{mv}_{\text {max }}}{\mathrm{eB}}

\text { or } r=\frac{\left(9.1 \times 10^{-31}\right) \times\left(1.3 \times 10^{\circ}\right)}{\left(5 \times 10^{-5}\right) \times\left(1.6 \times 10^{-10}\right)}=0.148 \text { meter . }

Posted by

Ritika Jonwal

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2024: NTA debars 39 candidates for 3 years on account of using unfair means
anam-khan

JEE Main 2024 session 2 paper 2 answer key objection window closes today; fees
anam-khan

JEE Main 2024 Topper: Farmer’s son Gajare Nilkrishna says ‘keep questioning until you understand’

Latest Question