Question

A gas of hydrogen-like ions is prepared in a particular excited state A. It emits photons having wavelength equal to the wavelength of the first line of the Lyman series together with photons of five other wavelengths. Identify the gas and find the principal quantum number of the state A.

Answer 1

So, the the gas is in $\textrm{4th excited state}$ and $\textrm{principal quantum number is 4}$ .

Explanation:

Given that,

Photons having wavelength equal to the wavelength of the first line of the Lyman series together with photons of five other wavelengths So there are total number of photons is 6.  

As we know that,

$\textrm{Number of photons}$  $= \frac{n(n-1)}{2}$  

$n^{2} - n -12 = 0$

$(n - 4)(n + 3) = 0$

$n = 4$

So, the the gas is in $\textrm{4th excited state}$ and $\textrm{principal quantum number is 4}$ .

Answer 2

Principal Quantum Number is 4

Explanation:

(a) If the atom is excited to the principal quantum (n), then the number of transitions is given by

$\frac {n(n-1)}{2}$

• It is given that a total of 6 photons are emitted. Therefore, total number of transitions is 6.

∴ $\frac {n(n-1)}{2} = 6$

⇒ n = 4

Thus, the principal quantum number is 4 and the gas is in the 4th excited state.