Threshold frequency, $v_0$ is the minimum frequency which a photon must possess to eject an electron from a metal. It is different for different metals. When a photon of frequency $1.0 \times 10^{15} \mathrm{~s}^{-1}$ was allowed to hit a metal surface, an electron having $1.988 \times 10^{-19} \mathrm{~J}$ of kinetic energy was emitted. Calculate the threshold frequency of this metal.
Show that an electron will not be emitted if a photon with a wavelength equal to 600 nm hits the metal surface.
$$\begin{aligned} &\begin{aligned} \text { We know that},\quad h \nu & =h \nu_0+\mathrm{KE} \\ \text { or } \quad h \nu-\mathrm{KE} & =h \nu_0=\left(6.626 \times 10^{-34} \mathrm{~J} \mathrm{~s} \times 1 \times 10^{15} \mathrm{~s}^{-1}\right)-1.988 \times 10^{-19} \mathrm{~J} \end{aligned} \end{aligned}$$
$$\begin{aligned} h \nu_0 & =6.626 \times 10^{-19}-1.988 \times 10^{-19} \mathrm{~J} \\ h \nu_0 & =4.638 \times 10^{-19} \mathrm{~J} \\ \nu_0 & =\frac{4.638 \times 10^{-19} \mathrm{~J}}{6.626 \times 10^{-34} \mathrm{Js}}=0.699 \times 10^{15} \mathrm{~s}^{-1} \end{aligned}$$
$$\begin{aligned} &\begin{aligned} \text { When, }\quad& \lambda=600 \mathrm{~nm}=600 \times 10^{-19} \mathrm{~m} \\ & \nu=\frac{c}{\lambda}=\frac{3.0 \times 10^8 \mathrm{~ms}^{-1}}{6.0 \times 10^{-7} \mathrm{~m}}=0.5 \times 10^{15} \mathrm{~s}^{-1} \end{aligned} \end{aligned}$$
$$\text { Thus, } \nu<\nu_0 \text {. hence, no electron will be emitted. }$$
When an electric discharge is passed through hydrogen gas, the hydrogen molecules dissociate to produce excited hydrogen atoms. These excited atoms emit electromagnetic radiation of discrete frequencies which can be given by the general formula
$$\bar{v}=109677\left[\frac{1}{n_i^2}-\frac{1}{n_f^2}\right]$$
What points of Bohr's model of an atom can be used to arrive at this formula? Based on these points derive the above formula giving description of each step and each term.
The two important points of Bohr's model that can be used to derive the given formula are as follows
(i) Electrons revolve around the nucleus in a circular path of fixed radius and energy. These paths are called orbits, stationary states or allowed energy states.
(ii) Energy is emitted or absorbed when an electron moves from higher stationary state to lower stationary state or from lower stationary state to higher stationary state respectively.
Derivation The energy of the electron in the $n^{\text {th }}$ stationary state is given by the expression,
$$E_n=-R_{\mathrm{H}}\left(\frac{1}{n^2}\right) \quad n=1,2,3\quad \text{... (i)}$$
where, $R_{\mathrm{H}}$ is called Rydberg constant and its value is $2.18 \times 10^{-18} \mathrm{~J}$. The energy of the lowest state, also called the ground state, is
$$E_n=-2.18 \times 10^{-18}\left(\frac{1}{1^2}\right)=-2.18 \times 10^{-18} \mathrm{~J}\quad \text{... (ii)}$$
The energy gap between the two orbits is given by the equation,
$$\Delta E=E_f-E_i,\quad \text{... (iii)}$$
On combining Eqs. (i) and (iii)
$$\Delta E=\left(-\frac{R_H}{n_f^2}\right)-\left(-\frac{R_H}{n_i^2}\right)$$
Where, $n_i$ and $n_f$ stand for initial orbit and final orbit.
$$\Delta E=R_{\mathrm{H}}\left[\frac{1}{n_i^2}-\frac{1}{n_f^2}\right]=2.18 \times 10^{-18} \mathrm{~J}\left[\frac{1}{n_i^2}-\frac{1}{n_f^2}\right]$$
Frequency, $\nu$ associated with the absorption and emission of the photon can be calculated as follows
$$\begin{aligned} & \nu=\frac{\Delta E}{h}=\frac{R_{\mathrm{H}}}{h}\left[\frac{1}{n_i^2}-\frac{1}{n_f^2}\right] \\ \Rightarrow \quad & \nu=\frac{2.18 \times 10^{-18} \mathrm{~J}}{6.626 \times 10^{-34} \mathrm{Js}}\left[\frac{1}{n_i^2}-\frac{1}{n_f^2}\right] \\ & \nu=3.29 \times 10^{15}\left[\frac{1}{n_i^2}-\frac{1}{n_f^2}\right] \mathrm{Hz} \\ \Rightarrow \quad & \bar{\nu}=\frac{\nu}{\mathrm{c}}=\frac{3.29 \times 10^{15}}{3 \times 10^8 \mathrm{~ms}^{-1}}\left[\frac{1}{n_i^2}-\frac{1}{n_f^2}\right] \\ & \bar{\nu}=1.09677 \times 10^7\left[\frac{1}{n_i^2}-\frac{1}{n_f^2}\right] \mathrm{m}^{-1} \\ & \bar{\nu}=109677\left[\frac{1}{n_i^2}-\frac{1}{n_f^2}\right] \mathrm{cm}^{-1} \end{aligned}$$
Calculate the energy and frequency of the radiation emitted when an electron jumps from $n=3$ to $n=2$ in a hydrogen atom.
In hydrogen spectrum, the spectral lines are expressed in term of wave number $\bar{v}$ obey the following formula
Wave number,
$$\begin{aligned} & \bar{\nu}=R_H\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) \quad\left(\text { where, } R_H=\text { Rydberg constant } 109677 \mathrm{~cm}^{-1}\right) \\ & \bar{\nu}=109677 \mathrm{~cm}^{-1}\left(\frac{1}{2^2}-\frac{1}{3^2}\right) \\ & \bar{\nu}=109677 \times \frac{5}{36}=15232.9 \mathrm{~cm}^{-1} \\ & \bar{\nu}=\frac{1}{\lambda} \end{aligned}$$
or, $$\lambda=\frac{1}{v}=\frac{1}{15232.9}=6.564 \times 10^{-5} \mathrm{~cm}$$
Wavelength, $$\lambda=6.564 \times 10^{-7} \mathrm{~m}$$
Energy,
$$\begin{aligned} E & =\frac{h c}{\lambda} \\ & =\frac{6.626 \times 10^{-34} \mathrm{Js} \times 3.0 \times 10^8 \mathrm{~ms}^{-1}}{6.564 \times 10^{-7} \mathrm{~m}} \\ & =3.028 \times 10^{-19} \mathrm{~J} \end{aligned}$$
Frequency,
$$\begin{aligned} \nu & =\frac{c}{\lambda}=\frac{3.0 \times 10^8 \mathrm{~ms}^{-1}}{6.564 \times 10^{-7} \mathrm{~m}} \\ & =0.457 \times 10^{15} \mathrm{~s}^{-1}=4.57 \times 10^{14} \mathrm{~s}^{-1} \end{aligned}$$
Why was a change in the Bohr Model of atom required? Due to which important development(s), concept of movement of an electron in an orbit was replaced by the concept of probability of finding electron in an orbital? What is the name given to the changed model of atom?
In Bohr model, an electron is regarded as a charged particle moving in well defined circular orbits about the nucleus. An orbit can completely be defined only if both the position and the velocity of the electron are known exactly at the same time. This is not possible according to the Heisenberg uncertainty principle. Further more, the wave character of the electron is not considered in Bohr model.
Therefore, concept of movement of an electron in an orbit was replaced by the concept of probability of finding electron in an orbital due to de-Broglie concept of dual nature of electron and Heisenberg's uncertainty principle. The changed model is called quantum mechanical model of the atom.