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) 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.

It is the minimum negative potential given to the anode in a photocell for which the photoelectric current becomes zero. If \({ v }_{ 0 }\) is the stopping potential, then maximum K.E. of emitted photoelectron is
\({ \left( K.E \right) }_{ max }={ eV }_{ 0 }=hv-{ \phi }_{ 0 }\)
\(V_{ 0 }=\frac { hv }{ e } -\frac { { \phi }_{ 0 } }{ e } \)
Threshold frequency  It is the minimum frequency of the incident radiation for which just emission of photoelectrons takes place from a metal surface without any K.E. If \(V_{ 0 }\)is the threshold frequency, then using Einstein's photoelectric equation
\(0={ hv }_{ 0 }-{ \phi }_{ 0 }\quad or\quad V_{ 0 }=\frac { { \phi }_{ 0 } }{ h } \)

De-Broglie wavelength associated with a body of mass m, moving with velocity v is given by
\(\lambda=\frac{h}{m v}\)
Since the mass of football is quite large, hence the de-Broglie wavelength associated with it is quite small hence it is not visible. 

The work function of a metal is the minimum energy required by an electron to just escape from the metal surface so as to overcome the restraining forces at the surface.
The relation between work function \(\left(\phi_0\right)\) and threshold frequency \(\left(v_0\right) \text { is } \phi_0=h v_0\) where h is Plank's constant. 

No, it may be absorbed in some other way. If the frequency of the incident photon is less than the threshold frequency, there will be no emission of photoelectrons at all.

The work function of alkali metal is very low. Even the ordinary visible light can bring about emission of photoelectrons from these metals. Due to this reason, alkali metal surfaces are most suited as photosensitive surfaces.

(a) Work function of caesium metal, ϕ₀ = 2.14eV
Frequency of light, v = 6.0 x 10¹⁴Hz
(a) The maximum kinetic energy is given by the photoelectric effect as:
= \(K=hv-{ \phi }_{ 0 }\)
Where,
h = Planck’s constant = 6.626 x 10⁻³⁴ Js
\(\therefore K=\frac{6.626 \times 10^{34} \times 6 \times 10^{14}}{1.6 \times 10^{-19}}-2.14\)
\(=2.485-2.140=0.345 \mathrm{eV}\)
Hence, the maximum kinetic energy of the emitted electrons is  0.345 eV.
(b) For stopping potential V₀, we can write the equation for kinetic energy as:
\(K=e V_{o}\)
\(\therefore V_{o}=\frac{K}{e}\)
\(=\frac{0.345 \times 1.6 \times 10^{-19}}{1.6 \times 10^{-19}}=0.345 \mathrm{~V}\)
Hence, the stopping potential of the material is 0.345 V.
c) Maximum speed of the emitted photoelectrons = v
Hence, the relation for kinetic energy can be written as:
\(K=\frac{1}{2} m v^{2}\)
Where,
m = Mass of an electron = 9.1 x 10⁻³¹ kg
\(v^{2}=\frac{2 K}{m}\)
\(=\frac{2 \times 0.345 \times 1.6 \times 10^{-19}}{9.1 \times 10^{-31}}=0.1104 \times 10^{12}\)
\(\therefore v=3.323 \times 10^{5} \mathrm{~m} / \mathrm{s}=332.3 \mathrm{~km} / \mathrm{s}\)
Hence, the maximum speed of the emitted photoelectrons is  332.3 km/s.

(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, \(\lambda=\frac{c}{v}=\frac{3 \times 10^8}{7.24 \times 10^{18}}=0.0414 \mathrm{~nm}\)

(i) (a)
(ii) (a): According to Einstein's photoelectric equation,
\(e V_{0}=\frac{h c}{\lambda}-\frac{h c}{\lambda_{0}}\)
As \(\lambda\)₀ is constant, so when \(\lambda\). is decreased, stopping potential (V₀) increases.
(iii) (b): It indicates that threshold frequency is greater than that of ultraviolet light. As X-rays have greater frequency than ultraviolet rays, so they can cause photoelectric effect.
(iv) (b): \(\mathrm{K.E}_{\cdot 1}=h \mathrm{v}-\phi\)
\({K.E}_{\cdot 2}=n h v-\phi=n(h v-\phi)+(n-1) \phi\)
\({K.E.} 2=n \mathrm{KE}_{1}+(n-1) \phi \)
\(\text { K.E. }_{2}>n \mathrm{KE}_{1}\)
(v) (b): Stopping potential is the measurement of maximum kinetic energy of emitted photoelectrons and kinetic energy of emitted photoelectrons is linearly related with the frequency of incident light corresponding (i.e., corresponding to shortest wavelength, KE. is maximum). Stopping potential is independent of intensity.