de-Broglie wavelength of an electron accelerated by a voltage of 50 V is close to

$(\left | e \right |=1.6\times 10^{-19}C,\: m_{e}=9.1\times 10^{-31}kg,\: h=6.6\times 10^{-34}Js)$

Option 1)
$0.5\dot{A}$
Option 2)
$1.2\dot{A}$

Option 3)
$1.7\dot{A}$
Option 4)
$2.4\dot{A}$

Answers (1)

P perimeter

As we learnt in

De - Broglie wavelength -

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

- wherein

$h= plank's\: constant$

$m= mass \: of\: particle$

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

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

${\lambda}=\frac{h}{p}=\frac{h}{\sqrt{2m.(K.E.)}}=\frac{h}{\sqrt{2m(eV)}}$

$=\frac{6.6\times10^{-34}}{\sqrt{2\times 9.1\times 10^{-31}\times{1.6}{\times10^{-19}{\times}50}}}m=1.7\AA$

Correct option is 3.

Option 1)

$0.5\dot{A}$

This is incorrect option.

Option 2)

$1.2\dot{A}$

This is incorrect option.

Option 3)

$1.7\dot{A}$

This is the correct option.

Option 4)

$2.4\dot{A}$

This is incorrect option.

Option 1) $0.5\dot{A}$	Option 2) $1.2\dot{A}$
Option 3) $1.7\dot{A}$	Option 4) $2.4\dot{A}$

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de-Broglie wavelength of an electron accelerated by a voltage of 50 V is close to Option 1) Option 2) Option 3) Option 4)

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de-Broglie wavelength of an electron accelerated by a voltage of 50 V is close to

$(\left | e \right |=1.6\times 10^{-19}C,\: m_{e}=9.1\times 10^{-31}kg,\: h=6.6\times 10^{-34}Js)$

Option 1)
$0.5\dot{A}$
Option 2)
$1.2\dot{A}$

Option 3)
$1.7\dot{A}$
Option 4)
$2.4\dot{A}$