# 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 emittedelectrons 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

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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