Compute the shortest and longest wavelengths in the Lyman series of hydrogen atoms.

Question

Vedantu Content Team · Accepted Answer

Hint: We need to use the wavelength formula given by Bohr model. The longest wavelength in Lyman series corresponds to the transition from first excited state to ground state while the shortest wavelength corresponds to a transition occurring from infinity to the ground state.Formula used:The formula for wavelength of Lyman series is given by the Bohr model as follows:$$\dfrac{1}{\lambda } = R\left( {\dfrac{1}{{{1^2}}} - \dfrac{1}{{{n^2}}}} \right)$$Complete answer:According to the Bohr model, the electron in a hydrogen atom can absorb energy and get excited to a higher energy state. It cannot stay in the higher energy state for too long and it undergoes de-excitation from the higher energy state to the lower energy state during which it loses its energy in the form of a photon.Lyman series are the series of wavelengths emitted by a hydrogen atom when an electron de-excites from an excited state to the ground state of the hydrogen atom. The value of the wavelength emitted by this particular de-excitation can be calculated by using the following formula.$$\dfrac{1}{\lambda } = R\left( {\dfrac{1}{{{1^2}}} - \dfrac{1}{{{n^2}}}} \right)$$                    …(i)Here n signifies the principal quantum number of the excited state and its value is $n > 1$ for the Lyman series. R is known as the Rydberg’s constant whose value is given as $R = 109737c{m^{ - 1}}$Now the shortest wavelength in these series corresponds to the de-excitation which occurs from infinitely high energy state to the ground state. For this, $n = \infty $. Using this in equation (i), we get$$  \dfrac{1}{\lambda } = R\left( {\dfrac{1}{{{1^2}}} - \dfrac{1}{{{\infty ^2}}}} \right) = R  \\   \Rightarrow \lambda  = \dfrac{1}{R} = \dfrac{1}{{109737}}cm = 9.11 \times {10^{ - 6}}cm = 91.1nm  \\  $$Next the longest wavelength in these series corresponds to a de-excitation which occurs from the first excited state to the ground state. For this, $n = 2$. Using this in equation (i), we get$$  \dfrac{1}{\lambda } = R\left( {\dfrac{1}{{{1^2}}} - \dfrac{1}{{{2^2}}}} \right) = R\left( {1 - \dfrac{1}{4}} \right) = \dfrac{{3R}}{4}  \\   \Rightarrow \lambda  = \dfrac{4}{{3R}} = \dfrac{4}{{3 \times 109737}}cm = 1.215 \times {10^{ - 5}}cm = 121.5nm  \\  $$These are the required values of wavelengths.Note: It should be noted that the energy emitted by a de-excitation is inversely proportional to the wavelength of the photon emitted. In case of the shortest wavelength, the energy emitted by the de-excitation is maximum while in case of the longest wavelength, the energy emitted by the de-excitation is minimum.

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