First slide

Photo Electric Effect

Question

In a photocell bichromatic light of wavelength 2475 $\overset{o}{A}$ and 6000 $\overset{o}{A}$ are incident on cathode whose work function is 4.8 eV. If a uniform magnetic field of $3 x 10^{- 5}$ Tesla exists parallel to the plate, the radius of the path describe by the photoelectron will be (mass of electron = $9 x 10^{- 31}$ kg)

Difficult

A

1 cm
B

5 cm
C

10 cm
D

25 cm

Solution

Energy of photons corresponding to light of wave
$\begin{array}{l} length λ_{1} = 2475 \overset{o}{A} is E_{1} = \frac{12375}{2475} = 5 eV \\ and that corresponding to λ_{2} = 6000 \overset{o}{A} is \\ E_{2} = \frac{12375}{6000} = 2 .06 eV \\ As E_{2} < W_{o} and E_{1} > W_{o} \\ Photoelectric emission is possible with λ_{1} only . \\ Maximum kinetic energy of emitted photoelectrons K = E - W_{o} = 5 - 4 .8 = 0 .28 eV . \\ Photo electrons experiences magnetic force and move along a circular path of radius \\ r = \frac{\sqrt{2 mk}}{QB} \frac{\sqrt{2 \times 9 \times 10^{- 31} \times 0 .2 \times 1 .6 \times 10^{- 19}}}{1 .6 \times 10^{- 19} \times 3 \times 10^{- 5}} = 0 .05 m = 5 cm \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App