Question

When a photosensitive surface is irradiated by light of wavelengths '$\lambda_1$' and '$\lambda_2$', kinetic energies of emitted photoelectrons are '$E_1$' and '$E_2$' respectively. The work function of the photosensitive surface is

• A. $\frac{\lambda_1 E_1 - \lambda_2 E_2}{\lambda_2 - \lambda_1}$
• B. $\frac{\lambda_1 E_1 + \lambda_2 E_2}{\lambda_2 - \lambda_1}$
• C. $\frac{\lambda_1 E_2 - \lambda_2 E_1}{\lambda_2 - \lambda_1}$
• D. $\frac{\lambda_1 E_2 + \lambda_2 E_1}{\lambda_2 - \lambda_1}$

Hint:

Multiply energy by its matching wavelength to isolate the constant $hc$: $\lambda E = hc - \lambda W$. Equating the two states and factoring out $W$ yields the correct algebraic ratio in just two quick lines of scratchpad layout!

Answer 1

The Correct Option is A

Step 1: Write Einstein's equation twice.
$E=\frac{hc}{\lambda}-W$ rearranges to $\lambda(E+W)=hc$. So $\lambda_1(E_1+W)=hc$ and $\lambda_2(E_2+W)=hc$.

Step 2: Set the two equal.
\[ \lambda_1(E_1+W)=\lambda_2(E_2+W) \]

Step 3: Collect the $W$ terms.
$\lambda_1E_1-\lambda_2E_2=W(\lambda_2-\lambda_1)$.

Step 4: Solve for $W$.
\[ W=\frac{\lambda_1E_1-\lambda_2E_2}{\lambda_2-\lambda_1} \] \[ \boxed{\frac{\lambda_1E_1-\lambda_2E_2}{\lambda_2-\lambda_1}} \]