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# What Transition in the Hydrogen Spectrum Would Have the Same Wavelength as the Balmer Transition N = 4 To N = 2 of He+ Spectrum - CBSE (Science) Class 11 - Chemistry

ConceptEvidence for the Quantized Electronic Energy Levels - Atomic Spectra

#### Question

What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition n = 4 to n = 2 of He+ spectrum?

#### Solution

For Heion, the wave number (barv)associated with the Balmer transition, n = 4 to = 2 is given by:

barv = 1/lambda =RZ^2(1/n_1^2 - 1/n_2^2)

Where

n1 = 2

n2 = 4

Z = atomic number of helium

barv = 1/lambda  = R(2)^2(1/4 - 1/16)

= 4R ((4-1)/16)

=barv = 1/lambda = (3R)/4

=> lambda = 4/(3R)

According to the question, the desired transition for hydrogen will have the same wavelength as that of He+.

=> R(1)^2[1/n_1^2 -1/n_2^2]  = (3R)/4

[1/n_1^2 - 1/n_2^2] = 3/4 .....(1)

By hit and trail method, the equality given by equation (1) is true only when

n1 = 1and n2 = 2.

∴ The transition for n2 = 2 to n = 1 in hydrogen spectrum would have the same wavelength as Balmer transition n = 4 to = 2 of He+ spectrum.

Is there an error in this question or solution?

#### APPEARS IN

Solution What Transition in the Hydrogen Spectrum Would Have the Same Wavelength as the Balmer Transition N = 4 To N = 2 of He+ Spectrum Concept: Evidence for the Quantized Electronic Energy Levels - Atomic Spectra.
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