Question

A small plate of a metal (work function $\varphi =1.17\text{eV})$ is placed at a distance of $2\text{m}$ from a monochromatic light source of wavelength $4800\mathring{A}$ and power $1.0\text{W}.$ The light falls normally on the plate. If a constant magnetic field of strength ${10}^{-4}\text{T}$ is applied parallel to the metal surface. Find the radius (in $\text{cm}$) of the largest circular path followed by the emitted photo electrons is (integer only)

[Take $hc=12431\mathring{A}\text{(eV)}$]

• A. 4.0
• B. 4
• C. 4.00

Answer 1

Answer: (A), (B), (C)

Given, $\varphi =1.17\text{eV};r=2\text{m}\lambda =4800\mathring{A};P=1.0\text{W}B={10}^{-4}\text{T}$

Now the energy of each photon is $E=\frac{hc}{\lambda }$

$E=\frac{12431}{4800}\text{eV}=2.58\text{eV}$

$=2.58\times 1.6\times {10}^{-19}=4.125\times {10}^{-19}\text{J}$

The rate of emission of photon per sec from the source is,

$r=\frac{P}{E}=\frac{\text{1}}{4.125\times {10}^{-19}}=2.424\times {10}^{18}$

No. of photon striking per square metal per sec on the plate is,

$n=\frac{2.425\times {10}^{18}}{4\times 0.314\times (2{)}^2}=4.82\times {10}^{16}$

The maximum kinetic energy of the photoelectrons emitted from the plate having work function $\varphi =1.17\text{eV}$ is given by,

$K{E}_{max}=E-\varphi$

$=2.58-1.17=1.41\text{eV}$

The maximum velocity of photo electrons ejected is gives as

$\frac{1}{2}m{v}_{max}^2=1.41\text{eV}$

$v_{max}=\sqrt{\frac{2\times 1.41\times 1.6\times {10}^{-19}}{9.1\times {10}^{-31}}}=7.036\times {10}^5{\text{ms}}^{-1}$

The radius of the circle traversed by photoelectrons in the magnetic field $B$ is given by,

$r=\frac{m{v}_{max}}{qB}=\frac{(9.1\times {10}^{-31})(7.036\times {10}^5)}{(1.6\times {10}^{-19})\left({10}^{-4}\right)}$

$=40.0\times {10}^{-3}\text{m}=4.0\text{cm}$