Question

In the H-spectrum, the longest wavelength of Lyman is 120 nm and shortest wavelength of Balmer from this data, find the longest wavelength of the photon that can ionize this H-atom. A proton and an electron, both at rest initially, combine to form an H-atom in a ground state. A is emitted in this process. Find the wavelength (in nm) of this photon.

Answer 1

Before solving this question, it is important to know the basic terms such as atomic number, mass number, neutrons, protons, electrons etc.

Complete answer:
Let’s understand the basic terminology:
Electrons, protons and neutrons are the subatomic particles. Electrons are the subatomic particle which carries a negative charge, while protons are the subatomic particle with a positive charge and neutrons are the subatomic particle which has no charge. Protons and neutrons both are present in the nucleus of the atom whereas electrons revolve around the nucleus. Given, A proton and an electron, both at rest initially, combine to form an H-atom in a ground state.

The hydrogen spectrum has the longest wavelength in lyman = 120nm

& \Rightarrow \dfrac{1}{\lambda }={{R}_{H}}(\dfrac{1}{{{1}^{2}}}-\dfrac{1}{{{\infty }^{2}}}) \\\ & \Rightarrow \lambda =\dfrac{1}{{{R}_{H}}}=120nm \\\ \end{aligned}$$ Shortest wavelength of Balmer series: $$\begin{aligned} &\Rightarrow \dfrac{1}{\lambda }={{R}_{H}}(\dfrac{1}{{{2}^{2}}}-\dfrac{1}{{{3}^{2}}}) \\\ & \Rightarrow \dfrac{1}{\lambda }={{R}_{H}}(\dfrac{1}{4}-\dfrac{1}{9}) \\\ & \Rightarrow \lambda =\dfrac{36}{5{{R}_{H}}}\times 120nm \\\ & \Rightarrow \lambda =864nm \\\ \end{aligned}$$ Hence, the energy emitted while forming the hydrogen bond = $$\begin{aligned} & \Rightarrow \dfrac{46}{\lambda }=2.176\times {{10}^{-18}} \\\ & \Rightarrow \lambda =\dfrac{46}{2.176\times {{10}^{-18}}} \\\ & \Rightarrow \lambda =\dfrac{6.626\times {{10}^{-34}}\times 3\times {{10}^{8}}}{2.176\times {{10}^{-18}}} \\\ & \Rightarrow \lambda =\dfrac{19.878}{2.176}\times {{10}^{-8}} \\\ &\Rightarrow \lambda =9.135\times {{10}^{-8}} \\\ \end{aligned}$$ **Hence, the wavelength of this photon in nm is 9.135$ \times {{10}^{-8}}$ .** **Note:** We can calculate frequency, wavelength and wavenumber if only one of the quantities is mentioned. Wavelength is the distance between the identical points in the adjacent cycles of the wave signals. And the wave number can be defined as the spatial frequency of a wave which is measured in cycles per unit distance or sometimes in radian per unit distance.