Question

Light of wavelength $4000A^o$ is incident on a metal surface of work function 2.5 eV. Given $h=6.62\times {10}^{-34}Js$, $c=3\times {10}^{8}m/s$, the maximum KE of photo-electrons emitted and the corresponding stopping potential are respectively

• A. 0. 6 eV, 0.6 V
• B. 2.5 eV, 2.5 V
• C. 3.1 eV, 3.1 V
• D. 0.6 eV, 0.3 V

Answer 1

The correct option is A 0. 6 eV, 0.6 V

According to given question, Maximum kinetic energy is K. E.,

$\frac{hC}{\lambda }-W=K.E.$
$\Rightarrow K.E.=4.14\times {10}^{-15}\times 3\times {10}^8-2.5$
$\Rightarrow K.E.=0.605eV$
$\Rightarrow K.E.\approx 0.6eV$
So,
From definition of stopping potential,
$K.E.=e\times V_0$ [Where $V_0$ is stopping potential]
$0.6eV=e{V}_0$
$\Rightarrow V_0=0.6volts$

So, the answer is option (A).