Question

Let $K_1$ be the maximum kinetic energy of photoelectrons emitted by a light of wavelength ${\lambda }_1$ and $K_2$ corresponding to ${\lambda }_1=2{\lambda }_2$ , then

• A. $2K_1-K_2$
• B. $K_1=2K_2$
• C. $K_1<\frac{K_2}{2}$
• D. $K_2>2k_1$

Answer 1

The correct option is D $K_2>2k_1$

By using Einstein’s photoelectric equation

$\frac{hc}{{\lambda }_1}=\varphi +K_1$

$\frac{hc}{2{\lambda }_2}=\varphi +K_2$

By dividing the second equation by$2$and subtracting the two equations we get

$\frac{hc}{2{\lambda }_2}-\frac{hc}{2{\lambda }_2}=K_1-\frac{k_2}{2}+\varphi -\frac{\varphi }{2}$

Hence we get

$\frac{K_2}{2}=K_1+\frac{\varphi }{2}$

Hence $K_2>2K_1$