Question

In a photo electric effect experiment, the maximum kinetic energy of the emitted electrons is $1eV$ for incoming radiation of frequency $v_0$ and $3eV$ for incoming radiation of frequency $3v_0/2$. What is the maximum kinetic energy of the electrons emitted for incoming radiations of frequency $9v_0/4$?

• A. $3eV$
• B. $4.5eV$
• C. $6eV$
• D. $9eV$

Answer 1

The correct option is C $6eV$

$(KE{)}_{max}=hv-{\varphi }_0$
So, $1eV=h{v}_0-{\varphi }_0....\left(i\right)$
and $3eV=\frac{h{v}_0}{2}-{\varphi }_0....\left(ii\right)$
$\Rightarrow 3eV-1eV=\frac{h{v}_0}{2}$
or $h{v}_0=4eV$
From Eq. (i), ${\varphi }_0=h{v}_0-1eV$
$=4eV-1eV=3eV$
$\therefore (KE{)}_{mas}=h\times \frac{9v_0}{4}-3eV$
$=\frac{9}{4}\left(4eV\right)-3eV=6eV$.