
What is the wavelength of the most energetic photon emitted in the Balmer series of the Hydrogen atom?
(a) $654 nm$
(b) $580 nm$
(c) $434 nm$
(d) $365 nm$
Answer
512.7k+ views
Hint: The Balmer series, or Balmer lines in nuclear physics, is one of six defined series explaining the spectral line radiations of the hydrogen atom. The shortest wavelength is the limit series, which is possessing high power.
Complete step-by-step solution:
Energy is related by wavelength by the formula:
$E = \dfrac{hc}{\lambda}$
$E \propto \dfrac{1}{\lambda}$
More energy, more will be the reverse of wavelength.
For Balmer Series of the hydrogen atom, we have-
$\dfrac{1}{\lambda} = R \left[ \dfrac{1}{2^{2}} - \dfrac{1}{n^{2}} \right]$
where, $n = 3, 4, 5…..$
Put $n = \infty$ in the above formula, we obtain the series limit of the Balmer series.
$\dfrac{1}{\lambda} = R \left[ \dfrac{1}{2^{2}} - \dfrac{1}{ \infty^{2}} \right]$
$\implies \dfrac{1}{\lambda} = R \left[ \dfrac{1}{2^{2}} - 0 \right]$
$\implies \lambda = \dfrac{4}{R}$
The value of $R$ for the hydrogen atom.
$R = 1.097 \times 10^{7} m^{-1}$
Put the value of R in the wavelength relation:
$ \lambda = \dfrac{4}{1.097 \times 10^{7} } m$
$\lambda = 364.6 \times 10^{-9} m$
Option (d) is correct.
Note:The Balmer series is instrumental in astronomy because the Balmer lines seem in various stellar objects due to the plenty of hydrogen in the universe and therefore are generally seen and relatively strongly related to lines from additional elements.
Complete step-by-step solution:
Energy is related by wavelength by the formula:
$E = \dfrac{hc}{\lambda}$
$E \propto \dfrac{1}{\lambda}$
More energy, more will be the reverse of wavelength.
For Balmer Series of the hydrogen atom, we have-
$\dfrac{1}{\lambda} = R \left[ \dfrac{1}{2^{2}} - \dfrac{1}{n^{2}} \right]$
where, $n = 3, 4, 5…..$
Put $n = \infty$ in the above formula, we obtain the series limit of the Balmer series.
$\dfrac{1}{\lambda} = R \left[ \dfrac{1}{2^{2}} - \dfrac{1}{ \infty^{2}} \right]$
$\implies \dfrac{1}{\lambda} = R \left[ \dfrac{1}{2^{2}} - 0 \right]$
$\implies \lambda = \dfrac{4}{R}$
The value of $R$ for the hydrogen atom.
$R = 1.097 \times 10^{7} m^{-1}$
Put the value of R in the wavelength relation:
$ \lambda = \dfrac{4}{1.097 \times 10^{7} } m$
$\lambda = 364.6 \times 10^{-9} m$
Option (d) is correct.
Note:The Balmer series is instrumental in astronomy because the Balmer lines seem in various stellar objects due to the plenty of hydrogen in the universe and therefore are generally seen and relatively strongly related to lines from additional elements.
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