Question

Calcium plate has maximum possible radiation of wavelength $\lambda$ of $400nm$ to eject electrons. Its work function is

• A. 2.3 eV
• B. 3.1 eV
• C. 4.5 eV
• D. None of these

Answer 1

Answer: (B)

The minimum energy required for the emission of photoelectron from a metal is called the work function $W$ of that metal. If $v_0$ is threshold frequency, then
$W=h{v}_0$
but frequency $\left(v_0\right)=\frac{\text{ velocity }(c)}{\text{ wavelength }(\lambda )}$
$\therefore W=\frac{hc}{\lambda }$
where, $h$ is Planck's constant and $c$ the speed of light.
Given, $h=6.6\times 10^{-34}Js,c=3\times 10^{8}m/s,$
$\lambda =400nm=400\times 10^{-9}m$
$W=\frac{6.6\times 10^{-34}\times 3\times 10^8}{400\times 10^{-9}}J$
$W=4.95\times 10^{-19}J$
$W=\frac{4.95\times 10^{-19}}{1.6\times 10^{-19}}=3.1eV$