Ncert Solutions Physics Class 12th

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

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

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

Planck’s constant, h = 6.626 $\times {10}^{-34}$ Js

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

The equation for De Bro

11.13 Kinetic energy of electron, $E_k=$ 120 eV = 120 $\times 1.6\times {10}^{-19}$ J

Planck’s constant, h = 6.626 $\times {10}^{-34}$ Js

Charge of an electron, e = 1.6 $\times {10}^{-19}$ C

Mass of electron, m = 9.1 $\times {10}^{-31}$ kg

The kinetic energy of electron can be written as $E_k=\frac{1}{2}$ m $v^2$

$v$ = $\sqrt{\frac{2E_k}{m}}$ = $\sqrt{\frac{2\times 120\mathrm{}\times 1.6\times {10}^{-19}}{9.1\mathrm{}\times {10}^{-31}}}$ = 6.496

11.12 Potential difference, V = 56V

Planck’s constant, h = 6.626 $\times {10}^{-34}$ Js

Charge of an electron, e = 1.6 $\times {10}^{-19}$ C

Mass of electron, m = 9.1 $\times {10}^{-31}$ kg

At equilibrium, the kinetic energy of each electron is equal to the acceleration potential. If v is the velocity of each electron, we can wri

11.11 Wavelength of the light produced by Argon laser, $\lambda$ = 488 nm = 488 $\times {10}^{-9}$ m

Stopping potential of the photoelectrons, $V_0$ = 0.38 V = $\frac{0.38}{1.6\times {10}^{-19}}$ $eV$

Planck’s constant, h = 6.626 $\times {10}^{-34}$ Js

Charge of an electron, e = 1.6 $\times {10}^{-19}$ C

Speed of the light, c = 3 $\times {10}^8$ m/s

From Ei

11.10 Frequency of the incident photon, $\nu =7.21\times {10}^{14}$ Hz

Maximum speed of electron, v = 6 $\times {10}^5$ m/s

Planck’s constant, h = 6.626 $\times {10}^{-34}$ Js

Mass of electron, m = 9.1 $\times {10}^{-31}$ kg

Let the threshold frequency = ${\nu }_0$

From the relation of threshold frequency and kinetic energy, we can write

$\frac{1}{2}$ m&nbs

11.9 Work function of the metal, ${\varnothing }_0$ = $4.2eV$

Charge of an electron, e = 1.6 $\times {10}^{-19}$ C

Planck’s constant, h = 6.626 $\times {10}^{-34}$ Js

Wavelength of the incident radiation, $\lambda$ = 330 nm= 330 $\times {10}^{-9}$ m

Speed of the light, c = 3 $\times {10}^8$ m/s

The energy of the incident photon is given as:

$E$ =&

11.8 Threshold frequency of the metal, ${\nu }_0=3.3\times {10}^{14}$ Hz

Frequency of the light incident on metal, $\nu =8.2\times {10}^{14}$ Hz

Charge of an electron, e = 1.6 $\times {10}^{-19}$ C

Planck’s constant, h = 6.626 $\times {10}^{-34}$ Js

Let the cut-off voltage for the photoelectric emission from the metal be $V_0$

The equation of the cut-off en

11.7 Power of the sodium lamp, P = 100 W

Wavelength of the emitted sodium light, $\lambda$ = 589 nm = 589 $\times {10}^{-9}$ m

Planck’s constant, h = 6.626 $\times {10}^{-34}$ Js

Speed of light, c = 3 $\times {10}^8$ m/s

Energy per photon associated with the sodium light is given as:

$E$ = $\frac{hc}{\lambda }$ = $\frac{6.626\mathrm{}\times {10}^{-34}\times 3\mathrm{}\times {10}^8}{589\mathrm{}\times {10}^{-9}}$ = 3.37&nb

11. The slope of the cut-off voltage (V) versus frequency ( $\nu )$ of an incident light is given as:

$\frac{\mathrm{V}}{\nu }=$ 4.12 $\times {10}^{-15}$ Vs

The relationship of V and $\nu$ is given as $h\nu$ = $eV$ or $\frac{\mathrm{V}}{\nu }=$ $\frac{h}{e}$

where e = Charge of an electron = 1.6 $\times {10}^{-19}$ C and h = Plank’s constant

Therefore,

11.5 The energy flux of sunlight, $?$ = 1.388 $\times {10}^3$ W/ $m^2$

Hence power of the sunlight per square meter, P = 1.388 $\times {10}^{3}W$

Speed of light, c = 3 $\times {10}^8$ m/s

Planck’s constant, h = 6.626 $\times {10}^{-34}$ Js

Average wavelength of photon, $\lambda$ = 550 nm = 550 $\times {10}^{-9}$ m

If n is the number of