Given, slope of graph, tan \(\theta\) = 4.12 \(\times\) 10^-15 V-s

Charge on electron, e = 1.6 \(\times\)10^-19 C
Slope of graph of cut off voltage versus frequency is tan \(\theta=\frac{V}{v}\)
We know that, bv = eV or \(\frac{V}{v}=\frac{b}{e}\)
\(\begin{aligned} \therefore & \frac{h}{e}=4.12 \times 10^{-15} \end{aligned}\)
\(\begin{aligned} \Rightarrow & h=1.6 \times 10^{-19} \times 4.12 \times 10^{-15} \end{aligned}\)
\(\begin{aligned} =6.592 \times 10^{-34} \mathrm{~J}-\mathrm{s} \end{aligned}\)

M2>2M1

23.6 MeV.

\(7.58\times { 10 }^{ -3 }a.m.u./N\)

\(9.23\times { 10 }^{ -16 }J\)

(1/3)\(\mathring { A } \)

\(3\times { 10 }^{ 4 }photons\)

Given, \(\phi_{0}=2.0 \mathrm{eV} ; h-6.63 \times 10^{-34} \mathrm{~J}-\mathrm{s}, \mathrm{KE}_{\max }-?\)
In one complete vibration twice the electric field becomes zero, so the frequency of incident light is given by
\(\mathrm{V}=\frac{1}{2} \times 2.4 \times 10^{15}=1.2 \times 10^{15} \mathrm{~Hz}\)
Hence, maximum kinetic energy,
\(\mathrm{KE}_{\max }=h v-\phi_{0}=\frac{6.63 \times 10^{-34} \times 1.2 \times 10^{15}}{1.6 \times 10^{-19}}-2=2.97 \mathrm{eV}\)

From law of conservation of momentum
\(\left| p \right| =\left| { p }_{ A } \right| +\left| { p }_{ B } \right| \)
or \(\frac { h }{ { \lambda }_{ c } } =\frac { h }{ \lambda _{ A } } +\frac { h }{ \lambda _{ B } } \)
or  \({ \lambda }_{ c }=\frac { { \lambda }_{ A }{ \lambda }_{ B } }{ { \lambda }_{ A }{ +\lambda }_{ B } } \)   
(i) If  \({ p }_{ A }>0,{ p }_{ B }>0\)
\({ \lambda }_{ c }=\frac { { \lambda }_{ A }{ \lambda }_{ B } }{ { \lambda }_{ A }+{ \lambda }_{ B } } \)
(ii) \({ p }_{ A }<0,{ p }_{ B }<0\)
\({ \lambda }_{ c }=\frac { { (-\lambda }_{ A })(-{ \lambda }_{ B }) }{ -{ \lambda }_{ A }-{ \lambda }_{ B } } \)
(iii) \({ p }_{ A }>0,{ p }_{ B }<0\)
\({ \lambda }_{ c }=\frac { { \lambda }_{ A }(-{ \lambda }_{ B }) }{ { \lambda }_{ A }-{ \lambda }_{ B } }\)
\(=\frac { { \lambda }_{ A }-{ \lambda }_{ B } }{ -{ \lambda }_{ B }-{ \lambda }_{ A } } \)
(iv) \({ p }_{ A }<0,{ p }_{ B }<0\)
\({ \lambda }_{ c }=\frac { { (\lambda }_{ A }){ \lambda }_{ B } }{ { \lambda }_{ A }+{ \lambda }_{ B } } \)
\(\\ =\frac { { \lambda }_{ A }{ \lambda }_{ B } }{ -{ \lambda }_{ A }-{ \lambda }_{ B } } \)

0.9 eV

Power of the sodium lamp, P = 100 W
Wavelength of the emitted sodium light, λ = 589 nm = 589 x 10⁻⁹ m
Planck’s constant, h = 6.626 x 10⁻³⁴ Js
Speed of light, c = 3 x 10⁸ m/s
(a)The energy per photon associated with the sodium light is given as:
\(E=h \frac{c}{\lambda}\)
\(=\frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{589 \times 10^{-9}}=3.37 \times 10^{-19} \mathrm{~J}\)
\(=\frac{3.37 \times 10^{-19}}{1.6} \times 10^{-19}=2.11 \mathrm{eV}\)
(b) Number of photons delivered to the sphere = n
The equation for power can be written as:
P = nE
\(\therefore n=\frac{P}{E}\)
\(=\frac{100}{3.37 \times 10^{-19}}=2.96 \times 10^{20} \text { photons/s }\)
Therefore, every second, 2.96 x 10²⁰ photons are delivered to the sphere.

(a) Quarks inside protons and neutrons carry fractional charges. This is because nuclear force increases extremely if they are pulled apart. Therefore, fractional charges may exist in nature; observable charges are still the integral multiple of an electrical charge.
(b) The basic relations for electric field and magnetic field are  \(\left(e V=\frac{1}{2} m v^{2}\right) \text { and }\left(e B v=\frac{m v^{2}}{r}\right)\)respectively.
These relations include e (electric charge), v (velocity), m (mass), V (potential), r(radius), and B (magnetic field). These relations give the value of velocity of an electron as \(\left(v=\sqrt{2 v\left(\frac{e}{m}\right)}\right. \text { and }\left(v=B r\left(\frac{e}{m}\right)\right)\) respectively.
(c) At atmospheric pressure, the ions of gases have no chance of reaching their respective electrons because of collision and recombination with other gas molecules. Hence, gases are insulators at atmospheric pressure. At low pressures, ions have a chance of reaching their respective electrodes and constitute a current. Hence, they conduct electricity at these pressures.
(d) The work function of a metal is the minimum energy required for a conduction electron to get out of the metal surface. All the electrons in an atom do not have the same energy level. When a ray having some photon energy is incident on a metal surface, the electrons come out from different levels with different energies. Hence, these emitted electrons show different energy distributions.
(e) The absolute value of energy of a particle is arbitrary within the additive constant. Hence, wavelength (λ) is significant, but the frequency (ν) associated with an electron has no direct physical significance. Therefore, the product νλ (phase speed) has no physical significance.
Group speed is given as:
\(V_{G}=\frac{d v}{d k} \)
\(=\frac{\mathrm{d} \mathrm{v}}{d\left(\frac{1}{\lambda}\right)}=\frac{\mathrm{d} \mathrm{E}}{\mathrm{dp}}=\frac{d\left(\frac{p^{2}}{2} m\right)}{\mathrm{dp}}=\frac{p}{m}\)
This quantity has a physical meaning.

(a) Power of the medium wave transmitter, P = 10 kW = 10⁴ W = 10⁴ J/s
Hence, energy emitted by the transmitter per second, E = 10⁴
Wavelength of the radio wave, λ = 500 m
The energy of the wave is given as:
\(E_{1}=\frac{\mathrm{hc}}{\lambda}\)
Where,
h = Planck’s constant = 6.6 x 10⁻³⁴ Js
c = Speed of light = 3 x 10⁸ m/s
\(\therefore E_{1}=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{500}=3.96 \times 10^{-28} \mathrm{~J}\)
= 3 x 10 ³¹
The energy (E₁) of a radio photon is very less, but the number of photons (n) emitted per second in a radio wave is very large.
The existence of a minimum quantum of energy can be ignored and the total energy of a radio wave can be treated as being continuous.
(b)Intensity of light perceived by the human eye, I = 10⁻¹⁰ W m⁻²
Area of a pupil, A = 0.4 cm² = 0.4 x 10⁻⁴ m²
Frequency of white light, ν= 6 x 10¹⁴ Hz
The energy emitted by a photon is given as:
E = hν
Where,
h = Planck’s constant = 6.6 x 10⁻³⁴ Js
∴ E = 6.6 x 10⁻³⁴ x 6 x 10¹⁴
= 3.96 x 10⁻¹⁹ J
Let n be the total number of photons falling per second, per unit area of the pupil.
The total energy per unit for n falling photons is given as:
E = n x 3.96 x 10⁻¹⁹ J s⁻¹ m⁻²
The energy per unit area per second is the intensity of light.
∴ E = I
n x 3.96 x 10⁻¹⁹ = 10⁻¹⁰
\(n=\frac{10^{-10}}{3.96 \times 10^{-19}}\)
= 2.52 x 10⁸ m² s⁻¹
The total number of photons entering the pupil per second is given as:
n_A = n x  A
= 2.52 x 10⁸ x 0.4 x 10⁻⁴
= 1.008 x 10⁴ s⁻¹
This number is not as large as the one found in problem (a), but it is large enough for the human eye to never see the individual photons.