Question 282: Dual Nature of Matter and Radiation

Question

Photoelectric emission is observed from a metallic surface for frequencies $v_1$ and $v_2$ of the incident light rays $\left(v_1>v_2\right)$. If the maximum values of kinetic energy of the photoelectrons emitted in the two cases are in the ratio of $1: \mathrm{k}$, then the threshold frequency of metallic surface is

Accepted Answer

Let the threshold frequency be $v_0$.
Step 1: Write the photoelectric equation
For incident frequency $v$, the maximum kinetic energy ($K_{max}$) of photoelectrons is:
$$ K_{max} = h(v - v_0) $$
where $h$ is PlanckÃ¢ÂÂs constant.
Step 2: Write equations for the two cases
For $v_1$:
$$ K_1 = h(v_1 - v_0) $$
For $v_2$:
$$ K_2 = h(v_2 - v_0) $$
Step 3: Use the ratio information
Given:
$$ \frac{K_1}{K_2} = \frac{1}{k} $$
So,
$$ \frac{h(v_1 - v_0)}{h(v_2 - v_0)} = \frac{1}{k} $$
(The $h$ cancels out)
$$ \frac{v_1 - v_0}{v_2 - v_0} = \frac{1}{k} $$
Step 4: Rearranging to solve for $v_0$
Cross-multiply:
$$ k(v_1 - v_0) = v_2 - v_0 $$
Expand and collect $v_0$ terms:
$$ k v_1 - k v_0 = v_2 - v_0 $$
Bring $v_0$ terms to one side:
$$ k v_1 - v_2 = k v_0 - v_0 $$
$$ k v_1 - v_2 = (k - 1) v_0 $$
$$ v_0 = \frac{k v_1 - v_2}{k - 1} $$
Step 5: Match with options
This matches
Option D
:
$$ \boxed{\frac{k v_1 - v_2}{k - 1}} $$

Answer

$\frac{\mathrm{k} \mathrm{v}_2-\mathrm{v}_1}{\mathrm{k}-1}$

Answer

$\frac{v_2-v_1}{k}$

Answer

$\frac{v_1-v_2}{k-1}$

Answer

$\frac{\mathrm{k} \mathrm{v}_1-\mathrm{v}_2}{\mathrm{k}-1}$

Dual Nature of Matter and Radiation

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