
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
513.3k+ 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.
Recently Updated Pages
Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Business Studies: Engaging Questions & Answers for Success

Trending doubts
Which are the Top 10 Largest Countries of the World?

What are the major means of transport Explain each class 12 social science CBSE

Draw a labelled sketch of the human eye class 12 physics CBSE

Why cannot DNA pass through cell membranes class 12 biology CBSE

Differentiate between insitu conservation and exsitu class 12 biology CBSE

Draw a neat and well labeled diagram of TS of ovary class 12 biology CBSE

