Question

Suppose, in certain conditions only those transitions are allowed to hydrogen atoms in which the principal quantum number n changes by 2. (a) Find the smallest wavelength emitted by hydrogen. (b) List the wavelength emitted by hydrogen in the visible range (380 nm to 780 nm).

Answer 1

Explanation:

It is given that the possible transitions are:

To n2 = 3 from n1 = 1

To n2 = 4 from n1 = 2

(a) Here, n2 = 3 and n1 = 1

Energy,$E=13.6\left(\frac{1}{n 12}-\frac{1}{n 22}\right)$

$E=13.6 \times \frac{8}{9}$     …..(1)

Energy E is $E=\frac{h c}{\lambda}$

where, h = Constant of Planck

c = light’s speed

λ = Radiation’s wavelength

Therefore, $E=6.63 \times 10-34 \times 108 \times \frac{9}{1300} \times 8$…..(2)

When the equations 1 and 2 are equated, we have

$\Rightarrow \lambda=0.027 \times 10^{-7}=103 \mathrm{nm}$

(b) In Balmer series, comes the visible radiation.

As there will be changes to ‘n’ by 2, we consider n = 2 to n = 4.

Energy,

$E 1=13.6 \times\left(\frac{1}{4}-\frac{1}{16}\right)=2.55 \text { eVIf }$

λ1 is the radiation’s wavelength, when there is a transition between quantum number to n = 4 and n = 2, then $255=\frac{1242}{\lambda 1}$

λ1 = 487 nm