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  • When photon of energy  strikes the surface of a metal A, the ejected photoelectrons have maximum kinetic energy 

When photon of energy 4.0eV strikes the surface of a metal A, the ejected photoelectrons have maximum kinetic energy T_A\: eV and de-Broglie wavelength \lambda _A. The maximum kinetic energy of photoelectrons liberated from another metal B by photon of energy 4.50eV is T_B = (T_A-1.5)eV. If the de-Broglie wavelength of these photoelectrons \lambda _B=2\lambda _A, then the work function(in eV) of metal B is: 
Option: 1 4
Option: 2 1.5
Option: 3 3
Option: 4 2

Answers (1)

best_answer

As we know \lambda =\frac{h}{\sqrt{2mT}}

where T=kinetic energy

So \frac{\lambda_A}{\lambda_B} = \sqrt{\frac{T_A}{T_B}}=\frac{1}{2}...(1)

Also Given \ \ T_B=T_A-1.5... (2)

Using equation (1) and (2)

T_A=2 \ eV , \ and \ T_B=1.5 \ eV

And using the Einstein equation

E_B=T_B+W_B\\ \Rightarrow W_B=E_B-T_B=(4.5-0.5)eV=4 \ eV

So the correct option is 1.

Posted by

vishal kumar

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