Question

When light of wavelength 2480 $A^0$ is incident on a metal surface electrons are emitted with a maximum KE of 2 eV. The maximum KE of photo-electrons, if light of wavelength 1240 $A^0$ is incident on the same surface would be

• A. 4eV
• B. 1 eV
• C. 2eV
• D. 7eV

Answer 1

Answer: (D)

7eV

From the 1st condition
$\frac{hC}{\lambda }-W=2eV$
$hintermsofeV=\frac{6.63\times {10}^{-34}}{1.6\times {10}^{-19}}$
$\Rightarrow \frac{4.14\times {10}^{-15}\times 3\times {10}^8\times {10}^{10}}{2480}$
$\Rightarrow Workfor=(5.012-2)eV$
$\Rightarrow W\approx 3eV$
So,
In 2nd case when wavelength of incident light is $1240A^{\circ }$,
$\Rightarrow \frac{hC}{\lambda }-W=K.E.$
$\Rightarrow K.E.=\frac{4.14\times {10}^{-15}\times 3\times {10}^{18}}{1240}-3$
$=10-3=7eV$

So, the answer is option (D).