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Standard XII

Physics

Line Spectra

For which tra...

Question

For which transition of $H e^{+}$ spectrum energy will be same as $2^{n d}$ line of Balmer series of H spectrum?

n₁=4 to n₂=8

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n₁=2 to n₂=4

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n₁=3 to n₂=4

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n₁=4 to n₂=5

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Solution

The correct option is A
n₁=4 to n₂=8

for $2^{n d}$ line in Balmer series of hydrogen atom $¯ ¯ ¯ v$ = $\frac{3 R}{16}$ ;

for helium ion = $¯ ¯ ¯ v$ = $4 \times R [\frac{1}{16} - \frac{1}{64}]$ = $\frac{4 R \times 3}{64}$ = $\frac{3 R}{16}$

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Similar questions

What electronic transition in $L i^{+ 2}$ produces the radiation of the same wave length as the first Line in the Lyman series of hydrogen?

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For which transition of He+ spectrum energy will be same as 2nd line of Balmer series of H spectrum?

The correct option is A n1=4 to n2=8 for 2nd line in Balmer series of hydrogen atom ¯¯¯v = 3R16; for helium ion = ¯¯¯v = 4 × R[116 − 164] = 4R × 364 = 3R16

For which transition of $H e^{+}$ spectrum energy will be same as $2^{n d}$ line of Balmer series of H spectrum?

The correct option is A
n₁=4 to n₂=8

for $2^{n d}$ line in Balmer series of hydrogen atom $¯ ¯ ¯ v$ = $\frac{3 R}{16}$ ;

for helium ion = $¯ ¯ ¯ v$ = $4 \times R [\frac{1}{16} - \frac{1}{64}]$ = $\frac{4 R \times 3}{64}$ = $\frac{3 R}{16}$