Question

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

Answer 1

"From the given

n = 4 and n = 2

$For\quad { He },\quad \frac { 1 }{ \lambda} \quad =\quad { R }_{ H }{ Z }^{ 2 }\left[ \frac { 1 }{ { 2 }^{ 2 } } \quad -\quad \frac { 1 }{ { 4 }^{ 2 } } \right]$

$For\quad H,\quad \frac { 1 }{ \lambda} \quad =\quad { R }_{ H }\left[ \frac { 1 }{ { { n }_{ 1 } }^{ 2 } } \quad -\quad \frac { 1 }{ { n }_{ 2 }^{ 2 } } \right]$

$\lambda$ is same

${ Z }^{ 2 }\left[ \frac { 1 }{ { 2 }^{ 2 } } \quad -\quad \frac { 1 }{ { 4}^{ 2 } } \right] \quad =\quad \left[ \frac { 1 }{ { { n }_{ 1 } }^{ 2 } } \quad -\quad \frac { 1 }{ { n }_{ 2 }^{ 2} } \right]$

$Z\quad =\quad 2$

$\left[ \frac { 1 }{ { 1 }^{ 2 } } \quad -\frac { 1 }{ { 2 }^{ 2 } } \right] \quad =\quad \left[ \frac { 1 }{ { { n }_{ 1 } }^{ 2 } } \quad -\frac { 1 }{ { n }_{ 2 }^{ 2 } } \right]$

${ n }_{ 1 }\quad =\quad 1\quad \quad and\quad { n }_{ 2 }\quad =\quad 2$"

Answer 2

Answer:

Hence, transition n2=2 to n1=1 will give spectrum of the same wavelength as that of Balmer transition, n2=4 to n1=2 in He+.