Question

Hydrogen atom in ground state is excited by a monochromatic radiation of lamda = 975

a. number of spectral lines in the resulting spectrum emitted will be

Answer 1

Given,
Wavelength , λ = 975 A°
so, energy associated with it , ∆E = 12400/λ (in A°) eV
∆E = 12400/975 = 12.7 eV

∵ ∆E = 13.6( 1/n₁² - 1/n₂²)
Here n₁ = 1 { because intial Hydrogen atom in ground state }
n₂ = n And ∆E = 12.7 eV

12.7 eV = 13.6 ( 1 - 1/n²)
⇒12.7/13.6 = 1 - 1/n²
⇒ (13.6 - 12.7)/13.6 = 1/n²
⇒0.9/13.6 = 1/n²
⇒ n² = 13.6/0.9 ≈ 16
⇒ n = 4

Hence, total number of spectral lines = n(n -1)/2
= 4 × (4 - 1)/2 = 4 × 3/2 = 6

∴ answer is 6

Answer 2

Hello Dear.

Here is the answer---


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Given---

   Wavelength(λ)  = 975 

   Using the Formula,
   
    Energy Associated = 12400/Wavelength 
                           ΔE = 12400/λ
                          ΔE =  12400/975
                         ΔE = 12.71 electron volt(eV)

Also, We know,
       
 Energy Associated = 13.6$\frac{1 }{ (n1)^{2} } - \frac{1}{ (n2)^{2} }$

 Where,

    ΔE = Energy Associated
          = 12.71 eV
 
     n1 = 1   
[Since the Hydrogen atom is in the ground State]


Thus, Putting these values in the Formulas,

         12.71 = 13.6 [1/(1)² - 1/(n₂)²]
         
   ⇒     n₂ = 13.6/0.89
   ⇒   (n₂)² = 15.28
   ⇒   n₂ = 3.9
          n₂ ≈ 4


For the Spectral Lines,

 Using the Formula,
        n(n - 1)/2
   = 4(4 - 1)/2
   = 4 × 3/2
   = 2 × 3
   = 6

Thus, the number of Spectral Lines in the Resulting Spectrum is 6.

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Hope it helps.

Have a Marvelous Day.