# Ultraviolet light of wavelength $2271\: A^{\circ}$ from a $100\: W$ mercury source irradiates a photocell made of molybdenum metal. If the stopping potential is $-1.3\: V$,will the photocell respond to a high intensity $(10^{-5}W/m^2)$ red light of wavelength $632.8\: A^{\circ}$ produced by a He-Ne laser?

$\begin {array} {1 1} (a)\;yes & \quad (b)\;no \\ (c)\;maybe & \quad (d)\;can’t\: be\: determined \end {array}$

Ans : (a)
$\lambda = 2271 \times 10^{-10} m\: \: \: \: \: \: \: V_{\circ}= 1.3 V$
So, $\phi_{\circ}=hv - eV_{\circ}$
$= \large\frac{hc}{\lambda} - eV_{\circ}$
$= \large\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{(2271 \times10^{-10} ) – 1.6 \times 10^{-19} \times 1.3}$
$= 8.72 \times 106{-19} – 2.08 \times 10^{-19}$
$= 6.64 \times 10^{-19} J$
$= \large\frac{6.64 \times 10^{-19}}{ 1.6 \times 10^{-19} } = 4.15\: eV$
So,$v_{\circ}$ = threshold frequency of metal = $\large\frac{\phi_{\circ}}{h}$
$= \large\frac{6.64 \times 10^{-19}}{6.6 \times 10^{-34} } = 1.006 \times 10^{15}Hz$
$\lambda_r=$ wavelength of red light $= 632.8\: A^{\circ} = 6328 \times 10^{-11}m$
So, $v_r= \large\frac{c}{\lambda_r} = \large\frac{3 \times 10^8}{ 6328 \times 10^{-11} } = 4.74 \times 10^{15}Hz$
Since,$v_{\circ} < v_r$ , so photocell would respond.

answered Dec 25, 2013
edited Mar 13, 2014