A metal exposed to light of wavelength $800\, nm$ and emits photoelectrons with a certain kinetic energy The maximum kinetic energy of photo-electron doubles when light of wavelength $500\, nm$ is used The work function of the metal is (Take $hc =1230 \,eV - nm )$

Question

Out of the following physical quantities which quantity has the same unit as that of Planck's constant?

Answer 1

To solve this problem, we need to apply the photoelectric effect equation, which is given by:

$E_k = \frac{hc}{\lambda} - \phi$

where:

From the problem, for the initial wavelength $\lambda_1 = 800 \, \text{nm}$, the kinetic energy is:

$E_{k1} = \frac{1230 \, \text{eV} \cdot \text{nm}}{800 \, \text{nm}} - \phi$

Simplifying:

$E_{k1} = \frac{1230}{800} - \phi = 1.5375 - \phi \, \text{eV}$

When the wavelength changes to $\lambda_2 = 500 \, \text{nm}$, the kinetic energy doubles:

$E_{k2} = 2E_{k1} = \frac{1230 \, \text{eV} \cdot \text{nm}}{500 \, \text{nm}} - \phi$

Solving this, we have:

$E_{k2} = \frac{1230}{500} - \phi = 2.46 - \phi \, \text{eV}$

Since $E_{k2} = 2E_{k1}$, we equate and simplify:

$2(1.5375 - \phi) = 2.46 - \phi$

$3.075 - 2\phi = 2.46 - \phi$

$\phi = 3.075 - 2.46 = 0.615 \, \text{eV}$

Thus, the work function of the metal is $0.615 \, \text{eV}$.

Therefore, the correct answer is $0.615 \, \text{eV}$.

The Correct Option is C