Question

An electromagnetic radiation of frequency $3\times {10}^{15}$ Hz falls on a photosurface whose work function is 4 eV. Then the maximum velocity of the photo electrons emitted from the surface is

(given $h=6.6\times {10}^{-34}Js$, $m=9\times {10}^{-31}$ Kg)

• A. $1.7\times {10}^{6}m{s}^{-1}$
• B. $3.4\times {10}^{6}m{s}^{-1}$
• C. $2.5\times {10}^{6}m{s}^{-1}$
• D. $2.0\times {10}^{6}m{s}^{-1}$

Answer 1

The correct option is A $1.7\times {10}^{6}m{s}^{-1}$

According to the question,
$\Rightarrow hv-W=E$
$\Rightarrow (6.6\times {10}^{-34}\times 3\times {10}^{15})-(4\times 1.6\times {10}^{-19})=\frac{1}{2}\times 9\times {10}^{-31}{v}^2$
We need to find $v$ from the above equation,
$\Rightarrow v^2=2.97778\times {10}^{12}$
$\Rightarrow v=1.72\times {10}^6$m/s
$\Rightarrow v\approx 1.7\times {10}^6$m/s

So, the answer is option (A).