Question

Photoelectrons are ejected from the surface of a metal having work function 4.5 eV. Then the impulse transmitted to the surface of the metal when the electron flies off due to collision of light quanta of energy 4.9eV.

• A. $3.93\times {10}^{-25}Kgm{s}^{-1}$
• B. $1.47\times {10}^{-25}Kgm{s}^{-1}$
• C. $3.43\times {10}^{-25}Kgm{s}^{-1}$
• D. $4.62\times {10}^{-25}Kgm{s}^{-1}$

Answer 1

The correct option is C $3.43\times {10}^{-25}Kgm{s}^{-1}$

Work function of metal $=4.5eV$
Energy of photons of light $=4.9eV$
Now,
Energy imparted to the electrons $=(4.9-4.5)eV$
$=0.4eV$
Now to find the velocity,

$\frac{1}{2}m{v}^2=0.4eV$

$\Rightarrow 9.1\times {10}^{-31}\times v^2=0.8\times 1.6\times {10}^{-19}$
$\Rightarrow v=3.750\times {10}^5$ m/s
So,
Impulse transmitted to surface of metal when an electron flies off = $mv$
$=9.1\times {10}^{-31}\times 3.750\times {10}^{5}kgm{s}^{-1}$
$\approx 3.412\times {10}^{-25}kgm{s}^{-1}$
$\approx 3.43\times {10}^{-25}kgm{s}^{-1}$

So, the answer is option (C).