In a photoelectric effect experiment the threshold wavelength of light is 380 nm. 

If the wavelength of incident light is 260 nm, the maximum kinetic energy of emitted

electrons will be:

Given  E (in eV) = \frac{1237}{\lambda (in\: \: nm)} 

  • Option 1)

    1.5 eV

     

  • Option 2)

     3.0 eV

  • Option 3)

    4.5 eV

  • Option 4)

    15.1eV

 

Answers (1)

 

Conservation of energy -

h\nu = \phi _{0}+\frac{1}{2}mv^{2}_{max}

h\nu = h\nu _{0}+\frac{1}{2}mv^{2}_{max}

 

h\left ( \nu-\nu _{0} \right )=\frac{1}{2}mv^{2}_{max}

where, h - Plank's constant\ \nu - Frequency\ \nu_{0} - threshold\ frequency\ \phi_{0}- work function

- wherein

 

 

\frac{hc}{\lambda}=\frac{hc}{\lambda_o}+(KE)_{max}

or

\frac{1237}{\lambda}=\frac{1237}{\lambda_o}+(KE)_{max}

So, 

(KE)_{max}=\frac{1237}{\lambda}-\frac{1237}{\lambda_o}

(KE)_{max}=1237(\frac{1}{\lambda}-\frac{1}{\lambda_o})

(KE)_{max}=1237(\frac{1}{260}-\frac{1}{380})

(KE)_{max}=1237\times \frac{120}{260\times 380}

(KE)_{max}=1.5eV


Option 1)

1.5 eV

 

Option 2)

 3.0 eV

Option 3)

4.5 eV

Option 4)

15.1eV

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