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A proton accelerated through a potential...

A proton accelerated through a potential difference of 100V, has de-Broglie wavelength `lamda_0` . The de-Broglie wavelength of an `alpha` -particle, accelerated through 800V is

A

`(lamda_0)/(sqrt2)`

B

`(lamda_0)/(2)`

C

`(lamda_0)/(4)`

D

`(lamda_0)/(8)`

Text Solution

Verified by Experts

The correct Answer is:
D
`lamda_p/lamda_alpha = sqrt( (m_alpha q_alpha V_alpha)/(m_p q_p V_p) ) = sqrt(( (4m_p) (2e) xx (800) )/( (m_p) xx (e ) xx (100) ) ) = 8`
`rArr lamda_alpha = (lamda_p)/(8) = (lamda_0)/(8)`
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