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  • De Broglie wavelength of an  particle accelerated through a potential difference of V is given by<div class='q

De Broglie wavelength of an \alpha  particle accelerated through a potential difference of V is given by

Option: 1

\frac{0.202}{\sqrt{V}}A^{\circ}

 

 

 


Option: 2

\frac{0.303}{\sqrt{V}}A^{\circ}


Option: 3

\frac{0.101}{\sqrt{V}}A^{\circ}


Option: 4

None of the above


Answers (1)

best_answer

As we learned

 

De - Broglie wavelength of Alpha particle -

\lambda _{\alpha -partical}= \frac{0.101}{\sqrt{v}}A^{\circ}

- wherein

A^{\circ}-Angstrom

v-potential \: difference

 

 

\lambda =\frac{h}{\sqrt{2mq\nu }}

h=6.62\times 10^{-34}J-S

m=6.68\times 10^{-27}kg

q=3.2\times 10^{-19c}

\lambda _{2}=\frac{0.101}{\sqrt{V}}A^{\circ}

Posted by

vinayak

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