Physics Ncert Solutions Class 12th

11.20 Potential difference across evacuated tube, V = 500 V

Specific charge of electron, e/m = 1.76 $*{10}^{11}$ C/kg

The speed of each emitted electron is given by the relation of kinetic energy as

$E_{kinetic}$ = $\frac{1}{2}m{v}^2$ = $eV$

$v=\sqrt{\frac{2eV}{m}}$ = $\sqrt{2V\frac{e}{m}}$ = $\sqrt{2*500*1.76*{10}^{11}}$ = 13.27 $*{10}^6$ m/s

Therefore, the speed of each emitted electron is 13.27 $*{10}^6$ m/s

Collector potential, V = 10 MV = 10 $*{10}^6$ V

The speed is given by $v=\sqrt{\frac{2eV}{m}}$ $=\sqrt{2V\frac{e}{m}}$ = $\sqrt{2*10\mathrm{}*{10}^6\mathrm{}*1.76*{10}^{11}}$ = 1.876 $*{10}^9$ m/s

This is not possible nothing can move faster than the light. In the above formula

$E_{kinetic}$ = $\frac{1}{2}m{v}^2$ can only be used in the non-relativistic limit, i.e. v << c

For very high speed problems, relativistic equations must be considered for solving them. In the relativistic limit, the total energy is given as :

$E=m{c}^2$ , where m = relativistic mass

$E=m_0\sqrt{(1-\frac{v^2}{c^2}})$ where $m_0$ = Mass of the particle at rest.

Hence, kinetic energy is given as

$E_{kinetic}$ = m $c^2$ - $m_0{c}^2$

11.19 Temperature of the Nitrogen molecule, T = 300 K

Atomic mass of nitrogen = 14.0076 u

Hence, mass of Nitrogen molecule, m = 2 $*$ 14.0076 u = 28.0152 u

We know, 1 u = 1.66 $*{10}^{-27}$ kg

So, m = 28.0152 $*$ 1.66 $*{10}^{-27}$ kg = 4.65 $*{10}^{-26}$ kg

Planck's constant, h = 6.626 $*{10}^{-34}$ Js

Boltzmann constant, k = 1.38 $*{10}^{-23}$ kg $m^2{s}^{-2}{K}^{-1}$

We have the expression that relates to mean kinetic energy ( $\frac{3}{2}kT)$ of the nitrogen molecule with root mean square speed ( $v_{rms}$ ) as:

$\frac{1}{2}m{v_{rms}}^2$ = $\frac{3}{2}kT$

$v_{rms}=\sqrt{\frac{3kT}{m}}$ = $\sqrt{\frac{3*1.38*{10}^{-23}*300}{4.65*{10}^{-26}}}$ = 516.814 m/s

De Broglie wavelength of the nitrogen molecule

$\lambda =\frac{h}{m{v}_{rms}}$ = $\frac{6.626\mathrm{}*{10}^{-34}}{4.65*{10}^{-26}*516.814}$ = 2.76 $*{10}^{-11}$ m = 0.0276 nm

Therefore, De Broglie wavelength is 0.0276 nm.

11.17 De Broglie wavelength of the neutron, $\lambda$ = 1.40 $*{10}^{-10}$ m

Mass of neutron, m = 1.66 $*{10}^{-27}$ kg

Planck's constant, h = 6.626 $*{10}^{-34}$ Js

Kinetic energy, $E_k$ = $\frac{1}{2}m{v}^2$ ………….(1)

De Broglie wavelength and velocity (v) are related as

$\lambda =\frac{h}{mv}$ …………….(2)

Combining equation (1) and (2), we get

$E_k$ = $\frac{1}{2}m({\frac{h}{m\lambda })}^2$ = $\frac{h^2}{2m{\lambda }^2}$ = $\frac{{(6.626\mathrm{}*{10}^{-34})}^2}{2*1.66\mathrm{}*{10}^{-27}*({1.40\mathrm{}*{10}^{-10})}^2}$ = 6.75 $*{10}^{-21}$ J = $\frac{6.75*{10}^{-21}}{1.6*{10}^{-19}}$ eV = 42.17 $*{10}^{-3}eV$

Temperature of the neutron, T = 300 K

Average kinetic energy of the neutron, $E_{k-avg}$ = $\frac{3}{2}kT$ ,

where k = Boltzmann constant = 1.38 $*{10}^{-23}$ kg $m^2{s}^{-2}{K}^{-1}$

$E_{k-avg}$ = $\frac{3}{2}kT$ = $\frac{3}{2}*1.38*{10}^{-23}*300=6.21*{10}^{-21}$ J

The relationship of De Broglie wavelength is given as:

$\lambda =\frac{h}{\sqrt{2*E_{k-avg}*m}}$ = $\frac{6.626\mathrm{}*{10}^{-34}}{\sqrt{2*6.21*{10}^{-21}*1.66\mathrm{}*{10}^{-27}}}$ = 1.46 $*{10}^{-10}$ m = 0.146 nm

11.16 Wavelength of electron, ${\lambda }_e$ = Wavelength of proton, ${\lambda }_p$ = 1.0 nm = 1 $*{10}^{-9}$ m

Planck's constant, h = 6.626 $*{10}^{-34}$ Js

From De Broglie wavelength relation, $\lambda =\frac{h}{p},$ where p = momentum

$p$ = $\frac{h}{\lambda }$ = $\frac{6.626\mathrm{}*{10}^{-34}}{1\mathrm{}*{10}^{-9}}$ = 6.626 $*{10}^{-25}$ kg.m/s. Since ${\lambda }_e={\lambda }_p$ , their momentum will be also equal.

The energy of photon is given by the relation: $E$ = $\frac{hc}{\lambda },$

where c = speed of light = 3 $*{10}^8$ m/s

$E$ = $\frac{6.626\mathrm{}*{10}^{-34}*3*{10}^8}{1\mathrm{}*{10}^{-9}}$ Js = $\frac{6.626\mathrm{}*{10}^{-34}*3*{10}^8}{1\mathrm{}*{10}^{-9}*1.6*{10}^{-19}}$ eV = 1242.38 eV = 1.242 keV

Kinetic energy of electron, having momentum p is given by the relation

$E_k$ = $\frac{1}{2}\frac{p^2}{m}$ where m = mass of electron = 9.1 $*{10}^{-31}$ kg.

Hence, $E_k$ = $\frac{1}{2}\frac{{(6.626\mathrm{}*{10}^{-25})}^2}{9.1\mathrm{}*{10}^{-31}}$ J = 2.412 $*{10}^{-19}$ J = $\frac{2.412*{10}^{-19}}{1.6*{10}^{-19}}$ eV = 1.51 eV

11.15 Mass of the bullet, m = 0.04 kg

Speed of the bullet, v = 1.0 km/s = 1000 m/s

Planck's constant, h = 6.626 $*{10}^{-34}$ Js

De Broglie wavelength of the bullet is given by the relation:

$\lambda =\frac{h}{m*v}$ = $\frac{6.626\mathrm{}*{10}^{-34}}{0.04*1000}$ = 1.65 $*{10}^{-35}$ m

Mass of the ball, m = 0.06 kg

Speed of the ball, v = 1.0 m/s

Planck's constant, h = 6.626 $*{10}^{-34}$ Js

De Broglie wavelength of the bullet is given by the relation:

$\lambda =\frac{h}{m*v}$ = $\frac{6.626\mathrm{}*{10}^{-34}}{0.06*1}$ = 1.10 $*{10}^{-32}$ m

Mass of the dust particle, m = 1.0 $*{10}^{-9}$ kg

Speed of the dust particle, v = 2.2 m/s

Planck's constant, h = 6.626 $*{10}^{-34}$ Js

De Broglie wavelength of the bullet is given by the relation:

$\lambda =\frac{h}{m*v}$ = $\frac{6.626\mathrm{}*{10}^{-34}}{1.0\mathrm{}*{10}^{-9}*2.2}$ = 3.01 $*{10}^{-25}$ m

11.14 (a) Wavelength of light of a sodium line, $\lambda$ = 589 nm = 589 $*{10}^{-9}$ m

Mass of electron, $m_e$ = 9.1 $*{10}^{-31}$ kg

Mass of a neutron, $m_n$ = 1.66 $*{10}^{-27}$ kg

Planck's constant, h = 6.626 $*{10}^{-34}$ Js

The kinetic energy of the electron $E_k=\frac{1}{2}{m}_e{v}^2$ ….(1)

The equation for De Broglie wavelength $\lambda =\frac{h}{m_e*v}$ or ${\lambda }^2=\frac{h^2}{{m_e}^2*v^2}$ ………(2)

$v^2=\frac{h^2}{{m_e}^2*{\lambda }^2}$