Ncert Solutions Physics Class 12th

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

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

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

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

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

The equation of the cut-off energy is given as:

$e{V}_0$ = $h(\nu -{\nu }_0)$ or

$V_0=\frac{h(\nu -{\nu }_0)}{e}$ = $\frac{6.626\mathrm{}*{10}^{-34}*(8.2*{10}^{14}-3.3*{10}^{14})}{1.6\mathrm{}*{10}^{-19}}$ V = 2.0292 V

Therefore, the cut-off voltage for the photoelectric emission is 2.0292V

11.7 Power of the sodium lamp, P = 100 W

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

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

Speed of light, c = 3 $*{10}^8$ m/s

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

$E$ = $\frac{hc}{\lambda }$ = $\frac{6.626\mathrm{}*{10}^{-34}*3\mathrm{}*{10}^8}{589\mathrm{}*{10}^{-9}}$ = 3.37 $*{10}^{-19}$ J = $\frac{3.37\mathrm{}*{10}^{-19}}{1.6*{10}^{-19}}$ eV = 2.11 eV

Let the number of photon delivered to the sphere = n

The equation of power can be written as $P=nE$

$n$ = $\frac{P}{E}$ = $\frac{100}{3.37\mathrm{}*{10}^{-19}}$ photons/sec = 2.97 $*{10}^{20}$ photons/s

Therefore, every second, 2.97 $*{10}^{20}$ are delivered to the sphere.

Therefore, every second, 2.97 $*{10}^{20}$ are delivered to the sphere.

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 $*{10}^{-15}$ 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 $*{10}^{-19}$ C and h = Plank's constant

Therefore, h = $e*\frac{\mathrm{V}}{\nu }$ = 1.6 $*{10}^{-19}$ $*$ 4.12 $*{10}^{-15}$ = 6.592 $*{10}^{-34}$ Js

Plank's constant = 6.592 $*{10}^{-34}$ Js

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

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

Speed of light, c = 3 $*{10}^8$ m/s

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

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

If n is the number of photon per square meter, incident on earth per second, the equation of power can be written as

$P=nE$ or $n$ = $\frac{P}{E}$

We know $E$ = $\frac{hc}{\lambda }$

Hence, $n$ = $\frac{P\lambda }{hc}$ = $\frac{1.388*{10}^3*550\mathrm{}*{10}^{-9}}{6.626\mathrm{}*{10}^{-34}*3\mathrm{}*{10}^8}$ = 3.84 $*{10}^{21}$ photons $m^2$ /s

Therefore, every second 3.84 $*{10}^{21}$ photons are incident per square meter on earth.

11.3 Photoelectric cut-off voltage,  $V_0$ = 1.5 V

Maximum kinetic energy of the photoelectrons emitted is given as

$K=e{V}_0$ , where e = charge of an electron = 1.6 $*{10}^{-19}$ C

K = 1.6 $*{10}^{-19}*1.5$ = 2.4 $*{10}^{-19}$ J

Hence, maximum kinetic energy of the photoelectrons emitted is 2.4 $*{10}^{-19}$.

6.17 Line charge per unit length = $\lambda$ = $\frac{Totalcharge}{Length}$ = $\frac{Q}{2\pi r}$

Where r = distance of the point within the wheel

Mass of the wheel = M

Radius of the wheel = R

Magnetic field, $\stackrel{?}{B}$ = $-B_0\stackrel{?}{k}$

At a distance r, the magnetic force is balanced by the centripetal force. i.e.

BQ $v$ = $\frac{M{v}^2}{r}$ , where v = linear velocity of the wheel. Then,

B $*2\lambda \pi r$ = $\frac{Mv}{r}$

v = $\frac{2B\lambda \pi r^2}{M}$

Angular velocity, $\omega =\frac{v}{R}$ = $\frac{2B\lambda \pi r^2}{MR}$

For r $\le a\le R,weget$ $\omega =-\frac{2B_0\lambda \pi a^2}{MR}\stackrel{?}{k}$

6.15 Length of the solenoid, l = 30 cm = 0.3 m

Area of cross-section, A = 25 ${cm}^2$ = 25 $*{10}^{-4}$ $m^2$

Number of turns on the solenoid, N = 500

Current in the solenoid, I = 2.5 A

Current flowing time, t = ${10}^{-3}$ s

We know average back emf, e = $\frac{d\varnothing }{dt}$ …………………. (1)

Where $d\varnothing =$ Change in flux = NAB ……………… .(2)

B = Magnetic field strength = ${\mu }_0\frac{NI}{l}$ ………………. (3)

Where ${\mu }_0$ = Permeability of free space = 4 $\pi *{10}^{-7}$ T m $A^{-1}$

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

e = $\frac{NAB}{dt}$ = $\frac{NA}{dt}*{\mu }_0\frac{NI}{l}$ = $\frac{{\mu }_0{N}^{2}AI}{lt}$ = $\frac{4\pi *{10}^{-7}*{500}^2*25\mathrm{}*{10}^{-4}*2.5}{0.3*{10}^{-3}}$ = 6.544 V

Hence the average back emf induced in the solenoid is 6.5 V