If the kinetic energy of a free electron doubles, its de Broglie wavelength changes by the factor

Option 1)

1/\sqrt{2}\;

Option 2)

\; \; \sqrt{2}\;

Option 3)

\; 1/2\;

Option 4)

\; \; 2

Answers (1)

As we learnt in

De - Broglie wavelength -

\lambda = \frac{h}{p}= \frac{h}{mv}= \frac{h}{\sqrt{2mE}}

- wherein

h= plank's\: constant

m= mass \: of\: particle

v= speed \: of\: the \: particle

E= Kinetic \: energy \: of \: particle

 

 {\lambda}=\frac{h}{p}=\frac{h}{\sqrt{2mE}}

If kinetic energy is double then {\lambda}  becomes \frac{1}{\sqrt{2}} times.

Correct answer is 1. 


Option 1)

1/\sqrt{2}\;

This is the correct option.

Option 2)

\; \; \sqrt{2}\;

This is an incorrect option.

Option 3)

\; 1/2\;

This is an incorrect option.

Option 4)

\; \; 2

This is an incorrect option.

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