The value of wavelength radiation emitted due to transition of electrons from n = 4 to n - 1 state in hydrogen atom will be

Question

The value of wavelength radiation emitted due to transition of electrons from n = 4 to n - 1 state in hydrogen atom will be (A) (5R/36) (B) (16/3R) (C) (36/5R) (D) (3R/16)

Tardigrade Admin · Accepted Answer

We know that, (1/λ)= RZ 2((1/ n 12)-(1/ n 22)) For hydrogen atom, atomic no. Z=1 So, (1/λ)= R ((1/2)2-(1/4)2)=(3 R /16) ⇒ λ=(16/3 R)

Q. The value of wavelength radiation emitted due to transition of electrons from $n$ = 4 to $n$ - 1 state in hydrogen atom will be

$\frac{5 R}{36}$

$\frac{16}{3 R}$

$\frac{36}{5 R}$

$\frac{3 R}{16}$

We know that, $\frac{1}{λ} = R Z^{2} (\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}})$
For hydrogen atom, atomic no. $Z = 1$
So, $\frac{1}{λ} = R (\frac{1}{2}^{2} - \frac{1}{4}^{2}) = \frac{3 R}{16}$
$< b r / >\Rightarrow λ = \frac{16}{3 R} < b r / >$

Q. The value of wavelength radiation emitted due to transition of electrons from n = 4 to n - 1 state in hydrogen atom will be

365R​

3R16​

5R36​

163R​