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Step-by-Step Solution
Step 1: Write down the known information
When light of frequency $n$ is incident on the metal surface, the maximum kinetic energy $K_{\max}$ of the emitted photoelectrons is:
$\frac{1}{2} m v^2 = h n - \phi,$
where:
$m$ is the mass of the electron.
$v$ is the maximum velocity of the emitted photoelectrons for frequency $n$.
$h$ is Planck’s constant.
$\phi$ is the work function of the metal.
Step 2: Express $hn$ in terms of $v$ and $\phi$
Rearrange the above equation to find $h n$:
$h n = \frac{1}{2} m v^2 + \phi.$
Step 3: Write the expression for the new frequency
When the frequency is increased to $3n$, the maximum kinetic energy of the emitted photoelectrons becomes:
$\frac{1}{2} m v_1^2 = h (3n) - \phi,$
where $v_1$ is the new maximum velocity for frequency $3n$.
Step 4: Substitute $hn$ from Step 2 into the expression for $3n$
Substitute $h n = \frac{1}{2} m v^2 + \phi$ into $h(3n)$:
$\frac{1}{2} m v_1^2 = 3 \bigl(\frac{1}{2} m v^2 + \phi \bigr) - \phi.$
Simplify this further:
$\frac{1}{2} m v_1^2 = \frac{3}{2} m v^2 + 3\phi - \phi = \frac{3}{2} m v^2 + 2\phi.$
Step 5: Compare the new velocity $v_1$ with $\sqrt{3} \, v$
Rewriting, we get:
$v_1^2 = 3 v^2 + \frac{4\phi}{m}.$
Since $\phi$ is a positive quantity (work function), the term $\frac{4\phi}{m}$ is positive. Thus:
$v_1^2 > 3 v^2 \quad \Longrightarrow \quad v_1 > \sqrt{3}\,v.$
Conclusion
The new velocity of the photoelectrons is greater than $\sqrt{3}\,v.$ Hence, the correct answer is: more than $\sqrt{3}\,v$.
