Question

Ultraviolet light of wavelength $300nm$ and intensity $1.0W/m^2$ falls on the surface of a photo-electric material. If one percent of the incident photons produce photoelectrons, then the number of photoelectrons emitted from an area of $1.0c{m}^2$ of the surface is nearly :

• A. 9.61 x ${10}^{14}$ per second
• B. 4.12 x ${10}^{13}$ per second
• C. 1.51 x ${10}^{12}$ per second
• D. 2.13 x ${10}^{11}$ per second

Answer 1

Answer: (C)

1.51 x ${10}^{12}$ per second

Number of photons falling on the surface $=\frac{300\times {10}^{-9}}{6.63\times 3}\times {10}^{34}\times {10}^{-8}$

$=1.50829\times {10}^{18}$

As only 1% of photo-electrons are liberated, so, only $1.50829\times {10}^{16}$ photons are ejected per second per $m^2$.

In an area of $1c{m}^2$ which is ${10}^{-4}$ of $m^2$.

So, in $1c{m}^2$ = $1.50829\times {10}^{16}\times {10}^{-4}$ photons of ejected per second

$\approx 1.51\times {10}^{12}$ per second.

So, the answer is option (C).