Assertion (A) It is impossible to determine the exact position and exact momentum of an electron simultaneously.
Reason (R) The path of an electron in an atom is clearly defined.
What is photoelectric effect? State the result of photoelectric effect experiment that could not be explained on the basis of laws of classical physics. Explain this effect on the basis of quantum theory of electromagnetic radiations.
Photoelectric effect When radiation with certain minimum frequency $\left(v_0\right)$ strike the surface of a metal, the electrons are ejected from the surface of the metal. This phenomenon is called photoelectric effect. The electrons emitted are called photoelectrons.
Equipment for studying the photoelectric effect. Light of a particular frequency strikes a clean metal surface inside a vacuum chamber. Electrons are ejected from the metal and are counted by a detector that measures their kinetic energy.
The result observed in this experiment were
(i) The electrons are ejected from the metal surface as soon as the beam of light strikes the surface, i.e., there is no time lag between the striking of light beam and the ejection of electrons from the metal surface.
(ii) The number of electrons ejected is proportional to the intensity or brightness of light.
(iii) For each metal, there is a characteristic minimum frequency, $v_0$ (also known as threshold frequency) below which photoelectric effect is not observed. At a frequency $v>v_0$, the ejected electrons come out with certain kinetic energy. The kinetic energies of these electrons increases with the increase of frequency of the light used.
The above observation cannot be explained by the electromagnetic wave theory. According to this theory, since radiations were continuous, therefore, it should be possible to accumulate energy on the surface of the metal, irrespective of its frequency and thus, radiations of all frequencies should be able to eject electrons.
Similarly, according to this theory, the energy of the electrons ejected should depend upon the intensity of the incident radiation.
Particle Nature of Electromagnetic Radiation
To explain the phenomena of 'black body radiation' and 'photoelectric effect', Max Planck in 1900, put forward a theory known after his name as Planck's quantum theory. This theory was further extended by Einstein in 1905.
The important points of this theory are as follows
(i) The radiant energy is emitted or absorbed not continuously but discontinuously in the form of small discrete packets of energy. Each such packet of energy is called a 'quantum'. In case of light, the quantum of energy is called a 'photon'.
(ii) The energy of each quantum is directly proportional to the frequency of the radiation, i.e., $$\quad E \propto v \text { or } E=h \nu$$
where, $h$ is a proportionality constant, called Planck's constant. Its value is approximately equal to $6.626 \times 10^{-27} \mathrm{erg}$ s or $6.626 \times 10^{-34} \mathrm{~J} \mathrm{~s}$.
(iii) The total amount of energy emitted or absorbed by a body will be some whole number quanta.
Hence, $E=n h \nu$ (where, $n$ is any integer).
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,
$$\begin{aligned} \nu & =\frac{c}{\lambda}=\frac{3.0 \times 10^8 \mathrm{~ms}^{-1}}{6.564 \times 10^{-7} \mathrm{~m}} \\ & =0.457 \times 10^{15} \mathrm{~s}^{-1}=4.57 \times 10^{14} \mathrm{~s}^{-1} \end{aligned}$$