Question

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

• A. $2K_1=K_2$
• B. $K_1=2K_2$
• C. $K_1<K_2/2$
• D. $K_1>2K_2$

Answer 1

Answer: (C)

Here,
$K_1=\frac{hc}{{\lambda }_1}-W$...(i)
and $K_2=\frac{hc}{{\lambda }_2}-W$...(ii)
Substituting ${\lambda }_1=2{\lambda }_2$ in Eq. (i), we get
$K_1=\frac{hc}{2{\lambda }_2}-W$
$K_1=\frac{1}{2}\left(\frac{hc}{{\lambda }_2}\right)-W$
$=\frac{1}{2}\left(K_2+W\right)-W$
$K_1=\frac{K_2}{2}-\frac{W}{2}$
or $K_1<\frac{K_2}{2}$