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When photons of energy 4.25 \mathrm{eV} strike the surface of a metal \mathrm{A}, the ejected photoelectrons have maximum kinetic energy, \mathrm{T_A} expressed in \mathrm{eV} and de-Broglie wavelength \lambda_A. The maximum kinetic energy of photoelectrons liberated from another metal \mathrm{B} by photons of energy \mathrm{4.70 \mathrm{eV}\: is\: T_B=\left(T_A-1.50 \mathrm{eV}\right)}. If the deBroglie wavelength of these photoelectrons is \mathrm{\lambda_B=2 \lambda_A}, then which of the following is incorrect:
 

Option: 1

the work function of \mathrm{\: A\: is\: 2.25 \mathrm{eV}}
 


Option: 2

the work function of \mathrm{B\: is \: 4.20 \mathrm{eV}}
 


Option: 3

\mathrm{T_A=2.00 \mathrm{eV}}
 


Option: 4

\mathrm{T_B=2.75 \mathrm{eV}}


Answers (1)

best_answer

\mathrm{ K_{\max }=E-W }

Therefore,

\mathrm{ T_A=4.25-W_A }            ....(1)

\mathrm{T_B=\left(T_A-1.50\right)=4.70-W_B }    ..........(2)

Eqs. (1) and (2) gives,

\mathrm{ W_B-W_A=1.95 \mathrm{eV} }                  .............(3)

de-Broglie wavelength is given by

\mathrm{ \lambda=\frac{h}{\sqrt{2 K m}} \: or \: \lambda \propto \frac{1}{\sqrt{K}} }where K=kinetic energy of the electron
\therefore \quad \frac{\lambda_B}{\lambda_A}=\sqrt{\frac{K_A}{K_B}}

\mathrm{or } \mathrm{\quad 2=\sqrt{\frac{T_A}{T_A-1.5}}}

This gives, \mathrm{T_A=2 \mathrm{eV}}

From Eq. (1) \mathrm{W_A=4.25-T_A=2.25 \mathrm{eV}}

or \begin{aligned} W_B & =4.20 \mathrm{eV} \\ \end{aligned}

\mathrm{T_B =4.70-W_B=4.70-4.20=0.50 \mathrm{eV}}

 

Posted by

Gaurav

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