A proton accelerated by potential difference 100 V have de-Broglie wavelength of `lambda_(0)`
`alpha`-particle is accelerated by same potential different.Its de-Broglie wavelength will be ……

Options

A. `2sqrt(2)lambda_(0)`
B. `(lambda_(0))/(2sqrt(2))`
C. `(lambda_(0))/(sqrt(2))`
D. `(lambda_(0))/(2)`

Answer

B

Explanation

`(1)/(2)mv^(2)`=eV
`therefore m^(2)v^(2)=p^(2)=2meV`
Now `lambda=(h)/(p)`
`therefore lambda=(h)/(sqrt(2meV))`
`therefore m^(2)v^(2)=p^(2)=2meV`
Now `lambda=(h)/(p)`
`therefore lambda=(h)/(sqrt(2meV))`
`therefore lambda prop (1)/(sqrt(me))`
`therefore (lambda_(alpha))/(lambda_(p))=sqrt((m_(p))/(m_(alpha)e_(alpha)))`
Now `m_(alpha)=4m_(p),e_(alpha)=2e_(p)`,
`therefore (lambda_(a))/(lambda_(p))=sqrt((m_(p)e_(p))/(4m_(p)xx2e_(p)))=(1)/(sqrt(8))`
`therefore (lambda_(alpha))/(lambda_(p))=(1)/(2sqrt(2))`
`therefore (lambda_(alpha))/(lambda_(0))=(1)/(2sqrt(2)) therefore lambda_(alpha)=(lambda_(0))/(2sqrt(2))`