de-Broglie wavelength of an electron in the nth Bohr orbit is `lambda_(n)` and the angular momentum is `J_(n)` then

Options

A. `J_n alpha lamda_n`
B. `lamda_n alpha 1//J_n`
C. `J_n alpha sqrtlamda_n`
D. None of these

Answer

A

Explanation

de-Broglie wavelength
`lamda = h/p = h/(mv) implies lamda alpha 1//v`
Now, v` alpha` Z/n or v `alpha ` 1/n
[By Bohr atom relation]
Hence, `lamda`n `alpha` n ... .. (i)
i.e., wavelength in nth orbit `lamda`n is
proportional to n (quantum number)
Now `J_n` = n`h/2pi`...(ii)
[ `J_n` = angular momentum of electron in nth orbit]
i.e. `J_n alpha` n
Hence, from Eqs. (i) and (ii),
`J_n alpha lamda_n`