In davision -Germer experiment electron emitted from filament is accelerated by V volt.Its de-Broglie wavelength will be……..

Options

A. `(sqrt(h))/(2Vem)`
B. `(h)/(sqrt(2Vem))`
C. `(sqrt(2Vem))/(h)`
D. `(2Vem)/(sqrt(h))`

Answer

B

Explanation

`ve=(1)/(2)mv^(2)`
`2Vem=m^(2)v^(2)`
2Vem=`p^(2) [because mv=p]`
`therefore =sqrt(2Vem)`
Now de-Broglie wavelength `lambda=(h)/(p)`
`therefore lambda=(h)/(sqrt(2Vem))`