A photon is emitted in transition from \( n = 4 \) to \( n = 1 \) level in a hydrogen atom. The corresponding wavelength for this transition is:
A photon is emitted in transition from \( n = 4 \) to \( n = 1 \) level in a hydrogen atom. The corresponding wavelength for this transition is:
Show Hint
For hydrogen atom transitions, use \[ E = 13.6 \left( 1 - \frac{1}{n^2} \right) \, \text{eV} \] and \[ \lambda = \frac{hc}{E} \] to find wavelength.
99.3 nm
941 nm
97.4 nm
94.1 nm
The Correct Option is D
Solution and Explanation
The energy difference during the transition from \( n = 4 \) to \( n = 1 \) is given by:
\[ E = h\nu = \frac{hc}{\lambda} \]
For a hydrogen atom, the energy difference can be expressed as:
\[ E = 13.6 \, \text{eV} \cdot \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right] \]
Here:
\[ n_1 = 1, \quad n_2 = 4 \]
Substitute the values:
\[ E = 13.6 \cdot \left[ 1 - \frac{1}{16} \right] = 13.6 \cdot \frac{15}{16} \]
\[ E = 12.75 \, \text{eV} \]
Now, using the relation \( E = \frac{hc}{\lambda} \):
\[ \lambda = \frac{hc}{E} \]
Given:
\[ h = 4 \times 10^{-15} \, \text{eVs}, \quad c = 3 \times 10^8 \, \text{m/s}, \quad E = 12.75 \, \text{eV} \]
Substitute the values:
\[ \lambda = \frac{(4 \times 10^{-15})(3 \times 10^8)}{12.75} \]
\[ \lambda = 94.1 \, \text{nm} \]
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Concepts Used:
Atomic Spectra
The emission spectrum of a chemical element or chemical compound is the spectrum of frequencies of electromagnetic radiation emitted due to an electron making a transition from a high energy state to a lower energy state. The photon energy of the emitted photon is equal to the energy difference between the two states.
Read More: Atomic Spectra
Spectral Series of Hydrogen Atom
Rydberg Formula:
The Rydberg formula is the mathematical formula to compute the wavelength of light.
\[\frac{1}{\lambda} = RZ^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})\]Where,
R is the Rydberg constant (1.09737*107 m-1)
Z is the atomic number
n is the upper energy level
n’ is the lower energy level
λ is the wavelength of light
Spectral series of single-electron atoms like hydrogen have Z = 1.
Uses of Atomic Spectroscopy:
- It is used for identifying the spectral lines of materials used in metallurgy.
- It is used in pharmaceutical industries to find the traces of materials used.
- It can be used to study multidimensional elements.