Question

Calculate the de-Broglie Wavelength of a particle accelerated
through a potential difference of 4000 volte. Given mass of
아
Proton = 1.6 7 X10-22 kq. Plank's constant = 6.62 x10134 Joule -sec​

Answer 1

${ \bold{ \underline{Given :}}}$

• Potential difference (Ve) = 4000 v

• Mass of proton (m) = $\sf \: 1.67 \times 10 {}^{ - 22} kg$

• Plank's constant (h) = $\sf \: 6.62 \times 10 {}^{-34} \:j/s$

$\bold{ \underline{To \: Find :}}$

• The de-Broglie wavelength of a particle.

$\bold{ \underline{Solution :}}$

The de Broglie wavelength formula is expressed as ;

$\sf \red \bigstar \: \lambda = \dfrac{h}{\sqrt{2mVe}}$

[ Put the values ]

$\sf \leadsto \: \lambda \: = \dfrac{6.62 \times 10 {}^{ - 34} }{ \sqrt{ 2 \times 1.67 \times 10 {}^{ - 22} \times 4000 } }$

$\sf \leadsto \: \lambda \: = \dfrac{6.62 \times 10 {}^{ - 34} }{ \sqrt{3.34 \times 10 {}^{ - 22} \times 4000 } }$

$\sf \leadsto \: \lambda \: = 6.62 \times 10 {}^{ - 34} \times 1.15 \times 10 {}^{ - 9}$

$\sf \leadsto \: \lambda \: = 7.613 \times 10 {}^{-43} m \: \green \bigstar$

Therefore,

The de broglie wavelength of particle is $\sf{7.613 \times 10^{-43}m.}$

Answer 2

$\huge{ \underline{ \underline \mathfrak \blue{Aƞswer - }}}$

Acceleration potential {V=100 v}V=100v .The de Broglie wavelength -:

$\lambda{is}λis$

$\lambda{h/p}λh/p =\frac{1.227}{\sqrt{V}}$

$\lambdaλ =\frac{1.227}{\sqrt{100}}$

nm={0.124nm}0.124nm .

• The de Broglie wavelenght associated with an electron in this case is of the order of X-ray wavelengths.