In hydrogen - like atom `(z = 1)` with line of Lyman series has wavelength `lambda` the de - broglie's wavelength of electron in the level from which it originated is also `lambda` Find the value of n ?

Answer

B, D

Explanation

nth line of lymam series mean electron jumping from `(n + 1) th` orbit to 1st orbit
for as electron to revolve in `(n + 1) th` orbit
`2m = (n+ 1) lambda`
`rArr lambda = (2 pi)/((n + 1)) xx r (2 pi)/((n + 1)) [ 0.529 xx 10^(-10)]((n + 1)^(2))/(Z)`
`rArr (1)/(lambda) = (Z)/(2 pi [0.529 xx 10^(-10)](n + 1))`
Also we know that when electron jumps `(n + 1)th` orbit to ist orbit
`(1)/(lambda) = RZ^(2)[(1)/(t^(2)) - (1)/((n + 1)^(2))] = 1.09 xx 10^(7)Z^(2)[1 - (1)/((n + 1)^(2))]`...(ii)
from(i) and (ii)
`(Z)/(2 pi (0.529 xx 10^(-10) ) ( n+ 1)) = 1.09 xx 10^(7)Z^(2) [(1 - (1)/((n + 1)^(2))]`
On solving we get `n = 24