Question

if the frequency of the incident light falling on a photosensitive material is doubled, then $K.E$. Of the emitted photoelectrons:

• A. remains constant
• B. becomes two times its initial value
• C. becomes more then two times its initial value
• D. becomes less then two times its initial value

Answer 1

The correct option is B becomes two times its initial value

Kinetic energy of photoelectron is $E_1$ and $E_2$ for incident light of frequency $\upsilon$ and $2\upsilon$ respectively

Now, $h\upsilon =E_1+{\varphi }_0.....\left(I\right)$

$2h\upsilon =E_2+{\varphi }_0.....\left(II\right)$

Now, put the value of $h\upsilon$ in equation (II)

$2(E_1+{\varphi }_o)=E_2+{\varphi }_0$

$E_2=2E_1+{\varphi }_0$

So, the kinetic energy of the emitted photoelectrons becomes more then two times its initial value