Question

In a sample of a hydrogen atoms, all the atoms are in 5th excited state. If they de...
the ground state, the ratio of longest and shortest wavelengths of emitted phot
(A) 6:1
(B) 875 : 11
(C) 384 : 9
(D) 35:1
CIT otom​

Answer 1

answer : option (B) 875 : 11

It is given that in a sample of a hydrogen atoms, all the atoms are in 5th excited state. if they degenerates the ground state.

the longest wavelength emitted photon, λ_(max),

here n₁ = 6 and n₂ = 5

1/λ_(max) = RZ²[1/6² - 1/5² ] = 11RZ²/900

the shortest wavelength emitted photon, λ_(min),

here n₂ = 6 and n₁ = 1

1/λ_(min) = RZ²[1/1² - 1/6²] = 35RZ²/36

now dividing equation (1) by equation (2) we get,

[1/λ_(max)]/[1/λ_(min)] = [11RZ²/900]/[35RZ²/36]

⇒λ_min/λ_max = 11/875

⇒λ_max/λ_min = 875/11

hence ratio of longest wavelength to shortest wavelength is 875 : 11.

Answer 2

We have to find, the ratio of longest and shortest wavelengths of emitted photon.

Solution:

We know that,

The longest wavelength emitted photon, $\lambda_{max}$

n₁ = 6 and n₂ = 5

$\dfrac{1}{\lambda_{max}}$ = R$Z^2$$(\dfrac{1}{5^2} -\dfrac{1}{6^2})$

= $\dfrac{11RZ^2}{900}$    

The shortest wavelength emitted photon, $\lambda_{mim}$

here n₂ = 6 and n₁ = 1

$\dfrac{1}{\lambda_{mim}}$ = R$Z^2$$(\dfrac{1}{1^2} -\dfrac{1}{6^2})$

= $\dfrac{35RZ^2}{36}$

∴ The ratio of longest and shortest wavelengths of emitted photon

$\dfrac{\dfrac{1}{\lambda_{max}}}{\dfrac{1}{\lambda_{mim}}}$ $=\dfrac{\dfrac{11RZ^2}{900}}{\dfrac{35RZ^2}{36}}$  

⇒$\dfrac{\lambda_{mim}}{\lambda_{max}}$ $=\dfrac{11}{875}$

⇒$\dfrac{\lambda_{max}}{\lambda_{min}}$ $=\dfrac{875}{11}$

⇒ $\lambda_{max}$ : $\lambda_{mim}$ = 875 : 11

∴ The ratio of longest and shortest wavelengths of emitted photon,

$\lambda_{max}$ : $\lambda_{mim}$ = 875 : 11

Thus, the required "option B)  875 : 11" is correct.