Question

Photoelectric emission is observed from a surface for frequencies $v_1$ and $v_2$ of the incident radiation ($v_1 > v_2$). If the maximum K.E. of the photoelectrons in two cases are in ratio 1 : K, then the threshold frequency $v_0$ is given by

• A. $\frac{v_2-v_1}{K-1}$
• B. $\frac{Kv_1-v_2}{K-1}$
• C. $\frac{Kv_2-v_1}{K-1}$
• D. $\frac{v_2-v_1}{K}$

Answer 1

Answer: (B)

$\frac{Kv_1-v_2}{K-1}$

Answer 2

The problem asks us to find the threshold frequency ($\nu_0$) for photoelectric emission, given two incident frequencies ($\nu_1$ and $\nu_2$) and the ratio of the maximum kinetic energies of the photoelectrons emitted in each case.

According to Einstein's photoelectric equation, the maximum kinetic energy ($K_{max}$) of a photoelectron is given by:

$K_{max} = h\nu - h\nu_0$

where $h$ is Planck's constant, $\nu$ is the frequency of the incident radiation, and $\nu_0$ is the threshold frequency.

For the first case, with incident frequency $\nu_1$:

$K_{max,1} = h\nu_1 - h\nu_0$ (Equation 1)

For the second case, with incident frequency $\nu_2$:

$K_{max,2} = h\nu_2 - h\nu_0$ (Equation 2)

We are given that the ratio of the maximum kinetic energies is $1:K$. This means:

$\frac{K_{max,1}}{K_{max,2}} = \frac{1}{K}$

Substitute the expressions for $K_{max,1}$ and $K_{max,2}$ from Equations 1 and 2 into this ratio:

$\frac{h\nu_1 - h\nu_0}{h\nu_2 - h\nu_0} = \frac{1}{K}$

Factor out $h$ from the numerator and denominator on the left side:

$\frac{h(\nu_1 - \nu_0)}{h(\nu_2 - \nu_0)} = \frac{1}{K}$

Cancel $h$:

$\frac{\nu_1 - \nu_0}{\nu_2 - \nu_0} = \frac{1}{K}$

Now, cross-multiply to solve for $\nu_0$:

$K(\nu_1 - \nu_0) = 1(\nu_2 - \nu_0)$

$K\nu_1 - K\nu_0 = \nu_2 - \nu_0$

Rearrange the terms to group $\nu_0$ terms on one side and other terms on the other side:

$K\nu_1 - \nu_2 = K\nu_0 - \nu_0$

$K\nu_1 - \nu_2 = \nu_0(K - 1)$

Finally, solve for $\nu_0$:

$\nu_0 = \frac{K\nu_1 - \nu_2}{K - 1}$