MCQ
When the electron in hydrogen atom jumps from fourth Bohr orbit to second Bohr orbit, one gets the
- ✓second line Balmer series
- Bfirst line of Balmer series
- Cfirst line of Pfund series
- Dsecond line of Paschen series.
✓
Answer
Correct option: A.
second line Balmer series
(a) : When the electron jumps from fourth Bohr orbit to second Bohr orbit in hydrogen atom, then the wavelength will be
$\lambda=\frac{364.56 m^2}{m^2-4}$, where $m=4$
$\Rightarrow \frac{364.56 \times 16}{16-4}=986.08 nm$
This wavelength corresponds to second line of Balmer series and $656.3 nm , 434.1 nm$ and $410.2 nm$ are the first, third and fourth line of Balmer series respectively.
$\lambda=\frac{364.56 m^2}{m^2-4}$, where $m=4$
$\Rightarrow \frac{364.56 \times 16}{16-4}=986.08 nm$
This wavelength corresponds to second line of Balmer series and $656.3 nm , 434.1 nm$ and $410.2 nm$ are the first, third and fourth line of Balmer series respectively.
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