Solution:

Ans. (i) Intensity of incident radiation $I=n h v$, where, $n$ is the number of photons incident per unit time per unitarea. For same intensity of two monochromatic radiations of frequency $v_{1}$ and $v_{2} .1 / 2$ a) $n_{1} h v_{1}=n_{2} h v_{2}$

$\begin{array}{ll}\text { As, } & v_{1}>v_{2} \ \Rightarrow & n_{2}>n_{1}\end{array}$

Therefore, the number of electrons emitted for monochromatic radiation of frequency $v_{2}$, will be more than that for radiation of frequency $v_{1}$. $1 / 2$ $h \nu=\phi_{0}+K_{\max }$

$\therefore$ For given $\phi_{0}$ (work function of metal),

$K_{\max }$ increases with $v$.

$\therefore$ The maximum kinetic energy of emitted photoelectrons will be more for monochromatic light of frequency $v_{1}\left(\right.$ as $\left.v_{1}>v_{2}\right)$.

[CBSE Marking Scheme, 2014]

Detailed Answer :

(i) According to the quantum theory number of photons per unit area in unit time is the intensity of radiation.

Hence, $\quad I=n h v$

where, $n=$ number of photons

$v=$ frequency of monochromatic radiation

Given

$$ \begin{aligned} I_{1} & =I_{2} \ n_{1} h v_{1} & =n_{2} h v_{2} \end{aligned} $$

$$ \begin{array}{ll} \text { if, } & v_{1}>v_{2} \ \text { then } & n_{1}<n_{2} \end{array} $$

Hence, number of photons having $v_{2}$ frequency of monochromatic radiation is more than the number of photons having $v_{1}$ frequency of monochromatic radiation.

(ii) Kinetic energy of the emitted photons is given by $K E=h v-\phi_{0} ;$ where, $\phi_{0}$ is the work function of photosensitive surface; which is same for both radiations, as it is characteristic of metal surface. $1 / 2$ So, Maximum kinetic energy of emitted photon is more for higher frequency of radiation. Hence kinetic energy of emitted photoelectrons are more with $v_{1}$ frequency of monochromatic radiation.