If `lamda_(0)` is the De-Broglie wavelength for a proton accelerated through a potential difference of 100 V, De- Broglie wavelength for prop- particle accelerated through the same potential difference is

Options

A. `2sqrt(2lamda_(0)`
B. `(lamda0)/2`
C. `(lamda0)/(2sqrt2)`
D. `(lamda0)/(sqrt(2`

Answer

C

Explanation

De- Broglie wavelength =`lambda_(o)`
Potential difference =100V
The expression for De - Broglie wavelength of `alpha`-particle
`lamda=(h)/sqrt(2m_(alpha" q"alphaV)`
`implieslamda=(h)/sqrt(2(4m_(p))(2q_(p))V`
`therefore`Mass of `alpha` particle =Four times of proton
and
Charge of `alpha` particle =Twice of proton )
`implieslamda=(1)/sqrt(8)(h)/sqrt(2m_(p)q_(p)V`
`lamda=(1)/sqrt(8)(lamda_(o))`
Therefore the DE- Brogli wavelength for `alpha` particle accelerated through the same potential difference is given as ,
`lamda=(1)/sqrt(8)(lamda_(o))`(`therefore` From equation (1))
`thereforelamda=(lamda_(o))/2sqrt2`