Given, cut-off voltage, V₀ = 1.5 V
Maximum kinetic energy is given by,
\(\begin{aligned} \mathrm{KE}_{\text {max }} & =e V_0=1.5 \space \mathrm{eV}=1.5 \times 1.6 \times 10^{-19} \end{aligned}\)
\(=2.4 \times 10^{-19} \mathrm{~J}\)

(a) Each photon has an energy
\(E=hv=6.63\times { 10 }^{ -34 }J s\times 6.0\times { 10 }^{ 14 }Hz\)
(b) If N is the number of photons emitted by the source per second, the power P transmitted in the beam equals N times the energy per photon E, so that P = N E. Then
\(n=\frac { P }{ E } =\frac { 2.0\times { 10 }^{ -3 }W }{ 3.98\times { 10 }^{ -19 }J } \)
\(=5.0\times { 10 }^{ 15 }\) photons per second.

(a) For the cut-off or threshold frequency, the energy h v₀ of the incident radiation must be equal to work function  Φ₀, so that
\({ V }_{ 0 }=\frac { { \phi }_{ 0 } }{ h } =\frac { 2.14eV }{ 6.63\times { 10 }^{ -34 }Js }\)
\(=\frac { 2.14\times 1.6\times { 10 }^{ -19 }J }{ 6.63\times { 10 }^{ -34 }Js } =5.16\times { 10 }^{ 14 }Hz\)
Thus, for frequencies less than this threshold frequency, no photoelectrons are ejected.
(b) Photocurrent reduces to zero, when maximum kinetic energy of the emitted photoelectrons equals the potential energy eV₀ by the retarding potential V₀. Einstein’s Photoelectric equation is 
\(e{ V }_{ 0 }=\frac { hc }{ \lambda } -{ \phi }_{ 0 }\)
\(or\ \lambda =\frac { hc }{ \left( e{ V }_{ 0 }+{ \phi }_{ 0 } \right) }\)
\(or \ \lambda =\frac { \left( 6.63\times { 10 }^{ -34 }Js \right) \times \left( 3\times { 10 }^{ 8 }m/s \right) }{ \left( e\times 0.6V+2.14eV \right) } \)
\(\lambda =\frac { 19.89\times { 10 }^{ -26 }Jm }{ 2.74\times 1.6\times { 10 }^{ -19 }J } =454nm\)

6 eV

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According to photon picture:
(i) Each quantum of radiation has energy hv
(ii) In photo - electric effect the electrons in the metal absorbs this quantum of energy (hv).
(iii) When this energy exceeds the minimum energy needed for the ejection of photoelectron, flow of photo current starts.

(a) For the electron:
Mass m = 9.11 x 10^–31 kg, speed v = 5.4 x 10⁶ m/s. Then, momentum
p = m v = 9.11 x 10^–31 (kg) x 5.4 x 10⁶ (m/s)
p = 4.92 x 10^–24 kg m/s
de Broglie wavelength,\(\lambda\) = h/p
\(=\frac{6.63 \times 10^{-34} \mathrm{Js}}{4.92 \times 10^{-24} \mathrm{~kg} \mathrm{~m} / \mathrm{s}}\)
\(\lambda=0.135 \mathrm{nm}\)
(b) For the ball:
Mass m’ = 0.150 kg, speed v’ = 30.0 m/s.
Then momentum p’ = m’ v ’ = 0.150 (kg) x 30.0 (m/s)
p ’= 4.50 kg m/s
de Broglie wavelength \(\lambda\)’ = h/p’.
\(=\frac{6.63 \times 10^{\pm 34} \mathrm{Js}}{4.50 \times \mathrm{kg} \mathrm{m} / \mathrm{s}}\)
\(\lambda^{\prime}=1.47 \times 10^{-34} \mathrm{~m}\)
The de Broglie wavelength of electron is comparable with X-ray wavelengths. However, for the ball it is about 10^–19 times the size of the proton, quite beyond experimental measurement.

The wavelength of a photon increases

Since, work function is given as,
\( W_0=h v_o \\ W_0 \propto v_0 \)
As work function of metal X is higher than metal Y, so metal X has higher threshold frequency than metal Y.

If v_o is the threshold frequency, then from photoelectric equation, we have
\({ ev }_{ 1 }=h{ v }_{ 1 }-{ \phi }_{ 0 } \ and \ { ev }_{ 2 }={ hv }_{ 2 }-{ \phi }_{ 0 }\)
\(e({ v }_{ 2 }-{ v }_{ 1 })=h({ v }_{ 2 }-{ v }_{ 1 })or \ h=\frac { e({ v }_{ 2 }-{ v }_{ 1 }) }{ ({ v }_{ 2 }-{ v }_{ 1 }) } \)
\(Now,{ ev }_{ 1 }={ hv }_{ 1 }-{ \phi }_{ 0 }={ hv }_{ 1 }-{ hv }_{ 0 }\)
\(or \ { v }_{ 0 }={ v }_{ 1 }-\frac { { ev }_{ 1 } }{ h } ={ v }_{ 1 }-{ ev }_{ 1 }\left[ \frac { { v }_{ 2 }-{ v }_{ 1 } }{ e({ v }_{ 2 }-{ v }_{ 1 }) } \right]\)
\(={ v }_{ 1 }-\frac { { v }_{ 1 }({ v }_{ 2 }-{ v }_{ 1 }) }{ ({ v }_{ 2 }-{ v }_{ 1 }) } =\frac { { v }_{ 1 }{ V }_{ 2 }-{ v }_{ 1 }{ V }_{ 1 }-{ v }_{ 2 }{ V }_{ 1 }+{ v }_{ 1 }{ V }_{ 1 } }{ ({ v }_{ 2 }-{ v }_{ 1 }) } \)

Yes; because the wavelength of ultraviolet light is less than the threshold wavelength 5200\(A˚\)

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De Broglie wavelength associated with He atom = 0.7268 x 10^-10m
Room temperature, T = 27°C = 27 + 273 = 300 K
Atmospheric pressure, P = 1 atm = 1.01 x 10⁵ Pa
Atomic weight of a He atom = 4
Avogadro’s number, N_A = 6.023 x 10²³
Boltzmann constant, k = 1.38 x 10⁻²³ J mol⁻¹ K⁻¹
Average energy of a gas at temperature T,is given as:
\(E=\frac{3}{2} k T\)
De Broglie wavelength is given by the relation:
\(\lambda=\frac{h}{\sqrt{2 m E}}\)
Where,
m = Mass of a He atom
\(=\frac{\text { Atomic weight }}{N_{A}} \)
\(=\frac{4}{6.023 \times 10^{23}} \)
\(=6.64 \times 10^{-24} g=6.64 \times 10^{-27} \mathrm{~kg} \)
\(\therefore \lambda=\frac{h}{\sqrt{3 m k T}} \)
\(=\frac{6.6 \times 10^{-34}}{\sqrt{3 \times 6.64 \times 10^{-27} \times 1.38 \times 10^{-23} \times 300}} \)
\(=0.7268 \times 10^{-10 \mathrm{~m}}\)|
We have the ideal gas formula:
PV = RT
PV = kNT
\(\frac{V}{N}=\frac{\mathrm{kT}}{P}\)
Where,
V = Volume of the gas
N = Number of moles of the gas
Mean separation between two atoms of the gas is given by the relation:
\(r=\left(\frac{V}{N}\right)^{\frac{1}{3}}=\left(\frac{\mathrm{kT}}{P}\right)^{\frac{1}{3}} \)
\(=\left[\frac{1.38 \times 10^{-23} \times 300}{1.01 \times 10^{5}}\right]^{\frac{1}{3}} \)
\(=3.35 \times 10^{-9} \mathrm{~m}\)
Hence, the mean separation between the atoms is much greater than the de Broglie wavelength.

The momentum of an electromagnetic wave of frequency v, wavelength \(\lambda\) is given by
\(p=\frac{b v}{c}=\frac{b}{\lambda} \text { or } \lambda=\frac{b}{p}\)
de-Broglie wavelength of photon, \(\lambda=\frac{b}{p}\)
Thus, wavelength of electromagnetic radiation is equal to the de-Broglie wavelength.

(i) Energy = eV= hv
or \(v=\frac{e V}{h}=\frac{1.6 \times 10^{-19} \times 30 \times 10^3}{6.63 \times 10^{-34}}\)
= 7.24 \(\times\) 10¹⁸ Hz
(ii) As, \(c=v \lambda\)
\(\therefore\) Wavelength, \(\lambda=\frac{c}{v}=\frac{3 \times 10^8}{7.24 \times 10^{18}}=0.0414 \mathrm{~nm}\)

(d) : \(\lambda=\frac{h}{\sqrt{2 m K}}, \text { i.e., } \lambda \propto \frac{1}{\sqrt{K}}\) 
\(\therefore \ \frac{\lambda^{\prime}}{\lambda}=\sqrt{\frac{K}{K^{\prime}}}=\sqrt{\frac{K}{4 K}}=\frac{1}{2} \text { or } \lambda^{\prime}=\frac{\lambda}{2}\)