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  • A particle moving with kinetic energy E has de Broglie wavelength . If energy 

A particle moving with kinetic energy E has de Broglie wavelength \lambda. If energy \Delta E is added to its energy, the wavelength become \frac{\lambda}{2}. Value of \Delta E, is:
Option: 1 2E
Option: 2 4E
Option: 3 3E
Option: 4 E

Answers (1)

best_answer

 

 

 

\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE}} \\\Rightarrow \frac{\lambda_2}{\lambda_1} = \frac{\frac{\lambda}{2}}{\lambda} = \frac{1}{2} \\\Rightarrow \frac{E_1}{E_2} = \frac{E}{E + \Delta E} \\\Rightarrow \frac{\lambda_2}{\lambda_1} = \sqrt{\frac{E_1}{E_2}} \\\Rightarrow \frac{1}{4} = \frac{E}{E + \Delta E} \\\Rightarrow \Delta E = 3E

Hence the option correct option is (3).  

Posted by

avinash.dongre

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