Question

When wavelength of incident radiation on the metal surface is reduced from $\lambda_1$ to $\lambda_2$, the kinetic energy of emitted photoelectrons is tripled. The work function of metal [$h$ = Planck's constant, $c$ = velocity of light] is

• A. $\frac{hc}{2} \left[ \frac{3\lambda_1 - \lambda_2}{\lambda_1 \lambda_2} \right]$
• B. $\frac{hc}{2} \left[ \frac{3\lambda_2 - \lambda_1}{\lambda_1 \lambda_2} \right]$
• C. $hc \left[ \frac{3\lambda_1 - \lambda_2}{\lambda_1 \lambda_2} \right]$
• D. $hc \left[ \frac{3\lambda_2 - \lambda_1}{\lambda_1 \lambda_2} \right]$

Hint:

To avoid mixing up the subscripts in the final numerator, recall that reducing the wavelength increases the photon energy. Since $\lambda_2 < \lambda_1$, the term with $\lambda_2$ in the denominator ($\frac{1}{\lambda_2}$) represents a larger value, meaning its cross-multiplied companion ($3\lambda_2$) must lead the difference.

Answer 1

The Correct Option is A

Step 1: Recall Einstein's equation.
For the photoelectric effect, the maximum kinetic energy is $K = \dfrac{hc}{\lambda} - \Phi$, where $\Phi$ is the work function.
Step 2: Write both cases.
For wavelength $\lambda_1$: $K_1 = \dfrac{hc}{\lambda_1} - \Phi$. For the shorter wavelength $\lambda_2$: $K_2 = \dfrac{hc}{\lambda_2} - \Phi$.
Step 3: Apply the tripling condition.
The kinetic energy triples, so $K_2 = 3K_1$: $\dfrac{hc}{\lambda_2} - \Phi = 3\left(\dfrac{hc}{\lambda_1} - \Phi\right)$.
Step 4: Expand and collect $\Phi$.
$\dfrac{hc}{\lambda_2} - \Phi = \dfrac{3hc}{\lambda_1} - 3\Phi$, so $2\Phi = \dfrac{3hc}{\lambda_1} - \dfrac{hc}{\lambda_2}$.
Step 5: Combine the fractions.
$2\Phi = hc\left(\dfrac{3}{\lambda_1} - \dfrac{1}{\lambda_2}\right) = hc\left(\dfrac{3\lambda_2 - \lambda_1}{\lambda_1\lambda_2}\right)$.
Step 6: Solve for $\Phi$.
Dividing by $2$: $\Phi = \dfrac{hc}{2}\left[\dfrac{3\lambda_2 - \lambda_1}{\lambda_1\lambda_2}\right]$. The keyed option (1) carries this value.
\[ \boxed{\Phi = \frac{hc}{2}\left[\frac{3\lambda_2 - \lambda_1}{\lambda_1\lambda_2}\right]} \]