# Photo emissivecell with exicting wavelength ? , the maximum kinetic energy of electron is K. If...

## Question:

Photo emissivecell with exicting wavelength {eq}\lambda{/eq}, the maximum kinetic energy of electron is K. If the excting wavelength is change to 3{eq}\lambda{/eq}/4 the kinetic energy of the fastest emitted electron will be.....

1. Equal to 3K/4

2. 4K/3

3. Less than 4K/3.

4. More than 4K/3.

## Photoemissive cell:

The photoemissive cell is the type of device that works on the application of light ray strikes on it. Photoemissive cells are used in light sensors and other detecting devices. These cells works only in the case when the light photon strikes of the surface of the cell.

Given data

• Initial Wavelength is, {eq}\lambda {/eq} .
• Final wavelength is, {eq}\dfrac{{3\lambda }}{4} {/eq} .
• Kinetic energy is, K .

The expression for the Einstein photoelectric equation due to initial wavelength is,

{eq}hc = \lambda \left( {K + \phi } \right)..................\left( 1 \right) {/eq}

Here, h is the Plank constant and c is the speed of light.

The expression for the Einstein photoelectric equation due to final wavelength is,

{eq}hc = \dfrac{{3\lambda }}{4}\left( {{K_f} + \phi } \right)..................\left( 2 \right) {/eq}

From equation (1) and (2).

{eq}\begin{align*} \lambda \left( {K + \phi } \right) &= \dfrac{{3\lambda }}{4}\left( {{K_f} + \phi } \right)\\ \left( {K + \phi } \right) &= \dfrac{3}{4}\left( {{K_f} + \phi } \right)\\ {K_f} &= \dfrac{{4\left( {K + \phi } \right)}}{3} - \phi \\ {K_f} &= \dfrac{{4K}}{3} + \dfrac{{4\phi }}{3} - \phi \end{align*} {/eq}

Here, {eq}\phi {/eq} is constant. Then, from above expression.

{eq}{K_f} = \dfrac{{4K}}{3} + C {/eq}

Thus, the final kinetic energy will be more than {eq}\dfrac{{4K}}{3} {/eq} .

Hence, the option (4) is correct.