If the linear momentum of a particle is `2.2xx10^(4)kg-ms^(-1)`, then what will be its de-broglie wavelength? (Take, `h=6.6xx10^(-34)`J-s)

Options

A. `3xx10^(-39)m`
B. `3xx10^(-29)nm`
C. `6xx10^(-29)m`
D. `6xx10^(-29)nm`

Answer

B

Explanation

The de-Broglie wavelength is given by `lamda=(h)/(mv)`
As from Bohr's theory, velocity of electron is given by
`vprop(1)/(n)`
So `lamda propn`
`therefore` For an electron at ground level (n=1) and at n=4 level
`(lamda_(1))/(lamda_(4))=(1)/(4) implies lamda_(4)=4lamda_(1)`