Question

Maximum kinetic energy of a photoelectron is $E$ when the wavelength of incident light is $\lambda$. If energy becomes four times when wavelength is reduced to one-third, then work-function of the metal is

• A. $\frac{3hc}{\lambda}$
• B. $\frac{hc}{3\lambda}$
• C. $\frac{hc}{\lambda}$
• D. $\frac{hc}{2\lambda}$

Answer 1

Answer: (B)

$\frac{hc}{3\lambda}$

Answer 2

Let the work function of the metal be $\phi$. According to Einstein's photoelectric equation, the maximum kinetic energy ($E_k$) of a photoelectron is given by:

$E_k = h\nu - \phi$
or
$E_k = \frac{hc}{\lambda} - \phi$

Case 1:
Maximum kinetic energy is $E$ when the wavelength of incident light is $\lambda$.
So,
$E = \frac{hc}{\lambda} - \phi \quad \text{(Equation 1)}$

Case 2:
The energy becomes four times ($4E$) when the wavelength is reduced to one-third ($\lambda/3$).
So,
$4E = \frac{hc}{(\lambda/3)} - \phi$
$4E = \frac{3hc}{\lambda} - \phi \quad \text{(Equation 2)}$

Now, we have a system of two equations. Substitute the expression for $E$ from Equation 1 into Equation 2:
$4 \left( \frac{hc}{\lambda} - \phi \right) = \frac{3hc}{\lambda} - \phi$

Expand the left side:
$\frac{4hc}{\lambda} - 4\phi = \frac{3hc}{\lambda} - \phi$

Rearrange the terms to solve for $\phi$:
$\frac{4hc}{\lambda} - \frac{3hc}{\lambda} = 4\phi - \phi$

Simplify both sides:
$\frac{hc}{\lambda} = 3\phi$

Finally, solve for $\phi$:
$\phi = \frac{hc}{3\lambda}$