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Question

A hydrogen atom is in ground state. In order to get six lines in its emission spectrum, wavelength of incident radiation should be

(Take: hc=12425 eV˙A)

A
800 ˙A
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B
825 ˙A
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C
975 ˙A
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D
1025 ˙A
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Solution

The correct option is C 975 ˙A
The number of possible emission lines is given by,

N=n(n1)2

So, To get six possible emission lines,

6=n2(n1)

12=n2n

n2n12=0

n24n+3n12=0

n(n4)+3(n4)=0

(n+3)(n4)=0

n=3,4

As, n should be positive, n=4

So, the electron has to get excited to third excited state to emit 6 emission lines.

Energy to be supplied to the hydrogen atom at the ground state to get excited to the n=4 is,

ΔE=E4E1

ΔE=0.85 eV(13.6 eV)

hcλ=12.75 eV

λ=12425 eV˙A12.75 eV

λ=975 ˙A

Hence, option (C) is correct.


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