Question

The kinetic energy of photoelectron emitted from the surface of a metal is \(7.2\times10^{-20}\,\text{J}\), when the metal is made to strike with light having wavelength \(\lambda\) nm. What is the value of \(\lambda\)? \[ (\text{Work function of metal}=4.5\,\text{eV};\; h=6.6\times10^{-34}\,\text{J s};\; c=3\times10^8\,\text{m s}^{-1};\; 1\,\text{eV}=1.6\times10^{-19}\,\text{J}) \]

• A. \(125\)
• B. \(250\)
• C. \(175\)
• D. \(150\)

Hint:

Use Einstein's photoelectric equation \[ \boxed{ \frac{hc}{\lambda}=\phi+K.E. } \] Always convert the work function from eV to joules before substituting.

Answer 1

The Correct Option is B

Step 1: Use Einstein's photoelectric equation. The photoelectric equation is \[ \frac{hc}{\lambda} = \phi+K.E. \] where \[ \phi=4.5\,\text{eV} =4.5\times1.6\times10^{-19} =7.2\times10^{-19}\,\text{J}. \] The kinetic energy is \[ K.E.=7.2\times10^{-20}\,\text{J}. \]

Step 2: Calculate the photon energy. Hence, \[ E = 7.2\times10^{-19} + 7.2\times10^{-20} = 7.92\times10^{-19}\,\text{J}. \] Now, \[ \lambda = \frac{hc}{E} = \frac{6.6\times10^{-34}\times3\times10^8} {7.92\times10^{-19}} = 2.5\times10^{-7}\,\text{m}. \]

Step 3: Convert into nanometres. Since \[ 1\,\text{nm}=10^{-9}\,\text{m}, \] \[ \lambda = 2.5\times10^{-7}\times10^9 = 250\,\text{nm}. \] Hence, \[ \boxed{\lambda=250\,\text{nm}.} \] Therefore, the correct option is \(\boxed{(B)}\).