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When photons of wavelength \lambda_1=2920 \AA strike the surface of metal A, the ejected photoelectron have maximum kinetic energy of  k_1 \mathrm{eV} and the smallest de-Broglie
wavelength of \lambda. When photons of wavelength \lambda_2=2640 \AA strike the surface of metal B the ejected photoelectrons have kinetic energy ranging from zero to k_2=\left(k_1-1.5\right) \mathrm{eV}. The smallest de-Broglie wavelength of electrons emitted from metal B is 2 \lambda .
k_1  Find ?

 

Option: 1

1


Option: 2

2


Option: 3

3

 


Option: 4

4


Answers (1)

Energy of photons incident on A is

E_1=\frac{h c}{\lambda_1}=\frac{12410}{2920}=4.25 \mathrm{eV}

k_1=4.25-\phi_1\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, ...................(i)

Energy of photons incident on B is

E_2=\frac{h c}{\lambda_2}=\frac{12410}{2640}=4.70 \mathrm{eV}

\therefore                                                   k_2=4.70-\phi_2

\therefore                                      k_1-1.5=4.70-\phi_2\, \, \, \, \, \, \, \, \, \, \, \, ...................(ii)

Combining (i) and (ii) gives 

\phi_2-\phi_1=1.95 \mathrm{eV}\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, ..................(iii)

The smallest de-Broglie wavelength corresponds to the electron with maximum kinetic energy

\therefore \quad \lambda=\frac{h}{\sqrt{2 m k_1}}

And 

2 \lambda=\frac{h}{\sqrt{2 m k_2}}

2=\sqrt{\frac{k_1}{k_2}}=\sqrt{\frac{k_1}{k_1-1.5}}

4=\frac{k_1}{k_1-1.5} \Rightarrow k_1=2 \mathrm{eV}

Posted by

Sumit Saini

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