Q: The total energy of an electron in the first excited state of the hydrogen atom is -3.4 eV. Find out its (i) kinetic energy and (ii) potential energy in this state.

Kinetic energy of electron is, \$KE= \dfrac{13.6 Z^2}{n^2} eV\$For the first excited state of the hydrogen atom, \$n=2\$ and \$Z=1\$\$\therefore KE=\dfrac{13.6}{2^2}=3.4 eV\$Total energy , \$E=KE+PE \Rightarrow -3.4=3.4+PE\$\$\therefore PE=-6.8\ eV\$

Q: A proton, a deuteron and an \$\alpha\$-particle accelerated through the same potential difference enter a region of uniform magnetic field, moving at right angles to \$B\$. What is the ratio of their kinetic energies?

Kinetic energy gained by charged particle of charge \$q\$ when accelerated under a potential difference \$V\$ will be \$KE=qV\$For a given \$V, KE\propto q\$For proton, deuteron and \$\alpha\$-particle the ratio of charges is \$1 : 1 : 2\$

Q: Using Rutherford model of the atom, derive the expression for the total energy of the electron in hydrogen atom. What is the significance of total negative energy possessed by the electron?OR Using Bohr's postulates of the atomic model derive the expression for radius of nth electron orbit. Hence obtain the expression for Bohrs radius.

According to Rutherford's model, we have\$\frac{mv^2}{r}=\frac{1}{4 \pi \epsilon_0} \frac{Ze^2}{r^2}\$\$\implies\$ \$=\frac{1}{4 \pi \epsilon_0}\frac{Ze^2}{r}\$Therefore,Total Energy \$= P.E+K.E\$\$T.E=-\frac{1}{4 \pi \epsilon_0}\frac{Ze^2}{r}+\frac12 mv^2\$\$=\frac12 \frac{1}{4 \pi \epsilon_0} \frac{Ze^2}{r}\$\$=-\frac{1}{8 \pi \epsilon_0} \frac{Ze^2}{r}\$Energy is negative implies that the electron- nucleus is a bound or attractive system. ORAccording to Bohr's atomic model, electrons revolve around the nucleus only in those orbits for which the angular momentum is an integral multiple of \$\frac{h}{2\pi}\$So, as per Bohr's postulate, we have\$mvr=\frac{nh}{2 \pi}\$ Therefore,\$\frac{m v^2}{r}= \frac{1}{4 \pi \epsilon_0}\frac{ Ze^2}{r^2}\$ \$mvr=\frac{nh}{2\pi}\$Therefore,\$m^2 v^2r^2 =\frac{n^2 h^2}{4 \pi^2}\$ This implies,\$r=\frac{\epsilon_0 n^2 h^2}{\pi Z e^2 m}\$ This is the required expression for Bohr's radius.

Q: What is the value of specific charge of an electron?

Specific Charge is defined as \${e}/{m}\$ For electron, \$e = 1.6 \times 10^{-19} C\$\$m = 9.1 \times 10^{-31} Kg \$So, \$e/m_e\$ = \$1.76 \times 10 ^{11}\$

Q: The angular momentum of an electron in Bohr's hydrogen atom whose energy is \$-3.4 eV\$, is

Energy of electron in nth orbit of hydrogen atom\$E_n = - \dfrac{13.6}{n^2} eV\$\$\Rightarrow 3.4 = - \dfrac{13.6}{n^2} \Rightarrow n^2 = 4\$or, \$n = 2\$Angular momentum of electron,\$L = \dfrac{nh}{2 \pi} = \dfrac{2h}{2 \pi} = \dfrac{h}{\pi}\$

Q: What is impact parameter in a scattering experiment?

The impact parameter is defined as the perpendicular distance between the path of a projectile and the center of a potential field created by an object that the projectile is approaching (see diagram). It is often referred to in nuclear physics (see Rutherford scattering) and in classical mechanics.

Q: A proton and an \$\alpha-\$particle are accelerated through the same potential. The ratio of their velocities is

Charge on proton \$=\alpha\$Charge on \$\alpha\$-particle = \$2\alpha\$Mass of proton \$= \beta\$Mass of \$\alpha\$-particle \$= 4\beta\$Kinetic energy of proton,\$\dfrac{1}{2} \times \beta V_{p}^{2} = \alpha V\$\$V_{p} = \sqrt{\dfrac{2\alpha V}{\beta}}\$And Kinetic energy of \$\alpha\$-particle,\$\dfrac{1}{2} \times 4 \beta \times V_{\alpha}^{2} = 2\alpha V\$\$\Rightarrow V_{\alpha} = \sqrt{\dfrac{\alpha V}{\beta}}\$So,Ratio of \$V_{p}: V_{\alpha} = \sqrt{2} : 1\$

Question 1

The total energy of an electron in the first excited state of the hydrogen atom is -3.4 eV. Find out its (i) kinetic energy and (ii) potential energy in this state.

Accepted Answer

Kinetic energy of electron is, $KE= \dfrac{13.6 Z^2}{n^2}   eV$For  the first excited state of the hydrogen atom, $n=2$ and $Z=1$$\therefore KE=\dfrac{13.6}{2^2}=3.4  eV$Total energy , $E=KE+PE \Rightarrow -3.4=3.4+PE$$\therefore PE=-6.8\  eV$

Question 2

A proton, a deuteron and an $\alpha$-particle accelerated through the same potential difference enter a region of uniform magnetic field, moving at right angles to $B$. What is the ratio of their kinetic energies?

Accepted Answer

Kinetic energy gained by charged particle of charge $q$ when accelerated under a potential difference $V$ will be $KE=qV$For a given $V,  KE\propto q$For proton, deuteron and $\alpha$-particle the ratio of charges is $1 : 1 : 2$

Question 3

Using Rutherford model of the atom, derive the expression for the total energy of the electron in hydrogen atom. What is the significance of total negative energy possessed by the electron?
OR
Using Bohr's postulates of the atomic model derive the expression for radius of nth electron orbit. Hence obtain the expression for Bohrs radius.

Accepted Answer

According to Rutherford's model, we have$\frac{mv^2}{r}=\frac{1}{4 \pi \epsilon_0} \frac{Ze^2}{r^2}$$\implies$ $=\frac{1}{4 \pi \epsilon_0}\frac{Ze^2}{r}$Therefore,Total Energy $= P.E+K.E$$T.E=-\frac{1}{4 \pi \epsilon_0}\frac{Ze^2}{r}+\frac12 mv^2$$=\frac12 \frac{1}{4 \pi \epsilon_0} \frac{Ze^2}{r}$$=-\frac{1}{8 \pi \epsilon_0} \frac{Ze^2}{r}$Energy is negative implies that the electron- nucleus is a bound or attractive system.                                                                                        ORAccording to Bohr's atomic model, electrons revolve around the nucleus only in those orbits for which the angular momentum is an integral multiple of $\frac{h}{2\pi}$So, as per Bohr's postulate, we have$mvr=\frac{nh}{2 \pi}$  Therefore,$\frac{m v^2}{r}= \frac{1}{4 \pi \epsilon_0}\frac{ Ze^2}{r^2}$     $mvr=\frac{nh}{2\pi}$Therefore,$m^2 v^2r^2 =\frac{n^2 h^2}{4 \pi^2}$     This implies,$r=\frac{\epsilon_0 n^2 h^2}{\pi Z e^2 m}$      This is the required expression for Bohr's radius.

Question 4

What is the value of specific charge of an electron?

Accepted Answer

Specific Charge is defined as ${e}/{m}$ For electron, $e = 1.6 \times 10^{-19} C$$m = 9.1 \times 10^{-31} Kg $So, $e/m_e$ = $1.76 \times 10 ^{11}$

Question 5

The angular momentum of an electron in Bohr's hydrogen atom whose energy is $-3.4 eV$, is

Accepted Answer

Energy of electron in nth orbit of hydrogen atom$E_n = - \dfrac{13.6}{n^2} eV$$\Rightarrow 3.4 = - \dfrac{13.6}{n^2} \Rightarrow n^2 = 4$or, $n = 2$Angular momentum of electron,$L = \dfrac{nh}{2 \pi} = \dfrac{2h}{2 \pi} = \dfrac{h}{\pi}$

Question 6

What is impact parameter in a scattering experiment?

Accepted Answer

The impact parameter is defined as&#10;the perpendicular distance between the path of a projectile and the center of a&#10;potential field created by an object that the projectile is approaching (see&#10;diagram). It is often referred to in nuclear physics (see Rutherford scattering) and in classical mechanics.

Question 7

The energy of the electron in the ground state of hydrogen atom is -13.6 eV. Find the kinetic energy and potential energy of electron in this state.

Accepted Answer

The kinetic energy, potential energy and total energy of an electron in hydrogen atom are related as-$KE=-TE$and $PE=2TE$Hence kinetic energy $KE=13.6eV$Potential energy $PE=-27.2eV$

Question 8

What were the three major observations Rutherford made in the gold foil experiment?

Accepted Answer

Three major observations Rutherford made in the gold foil experiment are1.Almost all the alpha particles did pass through the foil.2.Some alpha particles were deflected off at different angles as observed at the screen of the detector.3.Very few of the alpha particles (one or two) even bounced backwards after hitting the gold foil.

Question 9

A proton and an $\alpha-$particle are accelerated through the same potential. The ratio of their velocities is

Accepted Answer

Charge on proton $=\alpha$Charge on $\alpha$-particle = $2\alpha$Mass of proton $= \beta$Mass of $\alpha$-particle $= 4\beta$Kinetic energy of proton,$\dfrac{1}{2} \times \beta V_{p}^{2} = \alpha V$$V_{p} = \sqrt{\dfrac{2\alpha V}{\beta}}$And Kinetic energy of $\alpha$-particle,$\dfrac{1}{2} \times 4 \beta \times V_{\alpha}^{2} = 2\alpha V$$\Rightarrow V_{\alpha} = \sqrt{\dfrac{\alpha V}{\beta}}$So,Ratio of $V_{p}: V_{\alpha} = \sqrt{2} : 1$

Question 10

Define intensity of radiation on the basis of photon picture of light. Write its S.I. unit.

Accepted Answer

The intensity of radiation can be defined as the energy related with photons emitted from a unit surface area in unit time. Its S.I. units is watt per steradian. ($W{sr}^{-1}$)

Rutherford's atomic model

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