Question

Find the maximum $KE$ of photoelectrons emitted from the surface of lithium$(\varphi =2.39eV)$ when exposed with $E=E_0(1+\cos 6\times {10}^{14}t)\cos 3.6\times {10}^{15}t$

• A. 0.37 $eV$
• B. 0.1 $eV$
• C. 0.02 $eV$
• D. 0.06 $eV$

Answer 1

Answer: (C)

0.02 $eV$

For maximum $KE.$ frequency should be maximum. Here,
$w=3.6\times {10}^{15}{s}^{-1}$

$(KE{)}_{max}=\frac{hw}{2\pi }-\varphi$

$==\frac{6.625\times {10}^{-34}\times 3.6\times {10}^{+15}}{6.28\times 1.6\times {10}^{-19}}-2.39$

$=2.36-2.39$
$=0.02$ $eV$

So, the answer is option (C).