The kinetic energy acquired by the electron is given by 
\(=\frac{1}{2}mv^2=e V\)
Therefore, the speed v of the electron is \(v=\sqrt{\frac{2 eV}{m}}\)
Hence, the de Broglie wavelength of the electron is, \(\lambda=\frac{h}{mv}=\frac{h}{\sqrt{2meV}}\)

Photo cell is a device which converts light energy into electrical energy. It works on the principle of photo electric effect. When light is incident on photosensitive materials, their electric properties will get affected, based on which Photo cells are classified into three types. They are
(i) Photo emissive cell
(ii) Photo voltaic cell
(iii) Photo conductive cell

We know that \(\lambda=\frac{h}{\sqrt{2 m K}}\)
Since proton and electron have the same de Broglie wavelength, we get
\(\frac{h}{\sqrt{2 m_{p} K_{p}}}=\frac{h}{\sqrt{2 m_{c} K_{c}}} \text { (or) } \frac{K_{p}}{K_{e}}=\frac{m_{e}}{m_{p}}\)
Since m_e < m_p, K_e > K_p the electron has more kinetic energy than the proton.
\(\frac{K_{p}}{K_{e}}=\frac{\frac{1}{2} m_{p} v_{p}^{2}}{\frac{1}{2} m_{e} v_{e}^{2}} \text { (or) } \frac{v_{p}}{v_{e}}=\sqrt{\frac{K_{p} m_{e}}{K_{c} m_{p}}} \)
\(\frac{v_{p}}{v_{e}}=\sqrt{\frac{m_{e}^{2}}{m_{p}^{2}}}=\frac{m_{e}}{m_{p}} \text { since } \frac{K_{p}}{K_{e}}=\frac{m_{e}}{m_{p}} \)
Since m_e < m_p, v_p < v_e, the electron moves faster than the proton.

An alpha particle contains 2 protons and 2 neutrons. Therefore, the mass M of the alpha particle is 4 times that of a proton (m_p) (or a neutron) and its charge q is twice that of a proton (+e).
The de Broglie wavelength associated with it is
\(\lambda=\frac { h }{ \sqrt { 2MqV } } =\frac { h }{ \sqrt { 2\times (4m_{ p })\times (2e)\times V } } \)
\(=\frac { 6.626\times { 10 }^{ -34 } }{ \sqrt { 2\times 4\times 1.67\times 10^{ -27 }\times 2\times 1.6\times { 10 }^{ -19 }\times 400 } } \)
\(=\frac { 6.626\times 10^{ -34 } }{ 4\times 20\times { 10 }^{ -23 }\sqrt { 1.67\times 1.6 } } \) = 0.00507 \(\mathring { A }\)

i) Momentum of the electron is
p = \(\sqrt { 2mK } =\sqrt { 2\times 9.1\times { 10 }^{ -31 }\times 2\times 1.6\times 10^{ -19 } } \)
= 7.63 x 10^-25 kg ms^-1
Its de Broglie wavelength is
\(\lambda=\frac { h }{ p } =\frac { 6.626\times { 10 }^{ -34 } }{ 7.63\times { 10 }^{ -25 } } \) = 0.868 x 10^-9 m
= 8.68 \(\mathring { A }\)
ii) Momentum of the bullet is
p = m\({ \upsilon }\) = 0.050 x 200 = 10 kg ms^-1
It's de Broglie wavelength is
\(\lambda=\frac { h }{ p } =\frac { 6.626\times { 10 }^{ -34 } }{ 10 } \) = 6.626 x 10^-35 m
iii) Momentum of the car is
p = mv = 4000 x 50 = 2 x 10⁵ kg ms^-1
Its de Broglie wavelength is
\(\lambda=\frac { h }{ p } =\frac { 6.626\times { 10 }^{ -34 } }{ 2\times { 10 }^{ 5 } } \) = 3.313 x 10^-39 m
From these calculations, we notice that electron has a significant value of de Broglie wavelength (≈10^-9m which can be measured from diffraction studies) but the bullet and car have negligibly small de Broglie wavelengths associated with them (≈10^-33m and 10-³⁹m respectively, which are not measurable by any experiment). This implies that the wave nature of matter is important at the atomic level but it is not really relevant at the macroscopic level.

i) The work function is given by
ϕ₀ = hv - K_max = \(\frac { hc }{ \lambda } \) - eV₀
since K_max = eV₀
\(=\left[ \frac { 6.626\times { 10 }^{ -34 }\times 3\times { 10 }^{ 8 } }{ 390\times 10^{ -9 } } \right] \) - [1.6 x 10^-19 x 1.10]
= 5.10 x 10^-19 - 1.76 x 10^-19 = 3.34 x 10^-19 J
= 2.09 eV
ii) The threshold wavelength is
\(\lambda_{0}=\frac{h c}{\phi_o}=\frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{3.34 \times 10^{-19}}\) 
= 5.951 x 10^-7 m = 5951 \(\mathring { A }\).

i) The energy of the photon is
E = \(\frac { hc }{ \lambda } =\frac { 6.626\times { 10 }^{ -34 }\times 3\times { 10 }^{ 8 } }{ 3000\times { 10 }^{ -10 } } \)
E = 6.626 x 10^-19 J = 4.14 eV
Maximum KE of the photoelectrons is
K_max = hv - ϕ₀ = 4.14 - 2.30 = 1.84 eV
ii) The number of photons reaching the surface per second is
\(n_{p}=\frac{I}{E} \times A\) 
= \(\frac { 2 }{ 6.626\times 10^{ -19 } } \) x 2 x 10^-4
= 6.04 x 10¹⁴ photons / sec
The rate of emission of photoelectrons is
= (0.40) n_p = 0.4 x 6.04 x 10¹⁴
= 2.416 x 10¹⁴ photoelectrons/sec.

 \(\lambda_{p}=\frac{h}{\sqrt{2 m e V}} ; \lambda_{\alpha}=\frac{h}{\sqrt{m e V}} ; V=512 \mathrm{~V} \)
\(\frac{\lambda_{p}}{\lambda_{\alpha}}=\sqrt{\left(\frac{m_{\alpha}}{m_{p}}\right)\left(\frac{e_{\alpha}}{e_{p}}\right)\left(\frac{v_{\alpha }}{v_{p}}\right)} \)
\(\frac{m_{\propto}}{m_{p}}=4 ; \frac{e_{\alpha}}{e_{p}}=2 ; \frac{v_{\alpha}}{v_{p}}=\frac{x}{512} ; \frac{\lambda_{p}}{\lambda_{\alpha}}=1 \)
\(1=\sqrt{4 \times 2 \times\left(\frac{x}{512}\right)}=\frac{x}{64} \Rightarrow x=64 V \)

v = 81 V
\(\lambda=\frac{12.27}{\sqrt{\mathrm{V}}} \stackrel{o}A = \frac{12.27}{\sqrt{81}} \stackrel{o}A=1.363 \times 10^{-10} \mathrm{~m} \)
\(\lambda=1.363\stackrel{o}A\)
This wavelength falls in the region of X-ray spectrum

\(\text { (i) } \lambda_{\mathrm{d}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{meV_1}}} ; \lambda_{\alpha}=\frac{\mathrm{h}}{\sqrt{\mathrm{me}}}\)
\(\frac{\lambda_{\mathrm{d}}}{\lambda_{\alpha}}=\frac{\frac{1}{\sqrt{2 m e}}}{\frac{1}{\sqrt{4 m 2 e}}} \Rightarrow \lambda_{\mathrm{d}}=2 \lambda_{\alpha}\)
\(\text { (ii) } =\frac{K.E_d}{K.E_a}=\frac{eV}{2eV} =\frac{1}{2}\Rightarrow K.E_d=\frac{1}{2}K.E_a\)
∴ K.E of deuteron is half of K.E of α-particle.

\(\text { K.E }=81.9 \times 10^{-15} \mathrm{~J} \)
\(\lambda =\frac{\mathrm{h}}{\sqrt{2 \mathrm{mk}}}=\frac{6.626 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-3} \times 1836 \times 81.9 \times 10^{-15}}} \)
\(\lambda =\mathbf{4 . 0 0} \times 10^{-14} \mathrm{~m} \)

\(\lambda_{1} =1800 Å=1800 \times 10^{-10} \mathrm{~m} \)
\(\lambda_{2} =4965 Å=4965 \times 10^{-10} \mathrm{~m} \)
\(\mathrm{E} =\mathrm{hc}\left(\frac{1}{\lambda_{1}}-\frac{1}{\lambda_{2}}\right)=\frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{10^{-10}}\left(\frac{1}{1800}-\frac{1}{4965}\right) \)
\(E=\frac{7.04 \times 10^{-19}}{1.6 \times 10^{-19}}=4.399 \mathrm{eV} \simeq 4.4 \mathrm{eV} \)

\(\lambda=6000Å=6000 \times 10^{-10} \mathrm{~m} ; \mathrm{V}=0.8 \mathrm{v} \)
\(\mathrm{k} \cdot \mathrm{E}=\mathrm{hv}-\phi \)
\(\mathrm{eV}_o=\mathrm{hv}-\phi=\frac{\mathrm{hc}}{\lambda}-\phi \)
\((i) v=\frac{c}{\lambda}=\frac{3 \times 10^{8}}{6000 \times 10^{-10}}=5 \times 10^{14} \mathrm{~Hz} \)
\((ii)\ \mathrm{E}=\frac{\mathrm{hc}}{\lambda}=\frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{6000 \times 10^{-10}}=3.313 \times 10^{-19} J\)
\(\mathrm{E}=\frac{3.313 \times 10^{-19}}{1.6 \times 10^{-19}}=2.07 \mathrm{eV} \)
\((iii) \ \mathrm{E}=\mathrm{hv}-\mathrm{W}\Rightarrow\mathrm{W}=\mathrm{h} v-\mathrm{E} \)
\(\mathrm{E}=\mathrm{eV_o}=1.6 \times 10^{-19} \times 0.8=1.2 8 \times10^{-19}J\)
\(\mathrm{hv}=6.626 \times 10^{-34} \times 5 \times 10^{14}=3.313 \times 10^{-19}J \)
\(\mathrm{~W}=\frac{(3.313-1.28) \times 10^{-19}}{1.6 \times 10^{-19}}=1.270 \mathrm{eV} \)
W = 1.27 eV
\((iv) \ \mathrm{W}=\mathrm{h} \mathrm{v}_{0} \)
\(v_{0}=\frac{W}{h}=\frac{2.033 \times 10^{-19}}{6.626 \times 10^{-34}}=3.07 \times 10^{14} \mathrm{~Hz} \)
\((v)\ \mathrm{E}=\mathrm{eV}_o=\frac{0.8 \times 1.6 \times 10^{-19}}{1.6 \times 10^{-19}}=\mathbf{0 . 8} \mathrm{eV}\)

\(v= 9 \times 10^{14} \mathrm{~Hz} ; \mathrm{V}=8 \times 10^{5} \mathrm{~m} / \mathrm{s} \)
\(\mathrm{E}= \mathrm{h}-\mathrm{h} v_{o} \Rightarrow v_{o}=\frac{\mathrm{h}v-\mathrm{E}}{\mathrm{h}}=\mathrm{V}-\frac{\mathrm{E}}{\mathrm{h}}=\mathrm{V}-\frac{\frac{1}{2} \mathrm{mv}^{2} }{h}\)
\(v_s={9 \times 10^{14}-\frac{\left[\frac{1}{2} \times 9.1 \times 10^{-31} \times\left(8 \times 10^{5}\right)^{2}\right]}{6.626 \times 10^{-34}}}=4.605 \times 10^{14} \)
\( v_{o} \simeq 4.6 \times 10^{14} \mathrm{~Hz} \)

\(\lambda=4000 \stackrel{o}A=4000 \times 10^{-10} \mathrm{~m} \)
\(\lambda=\frac{\mathrm{h}}{\mathrm{mv}}\Rightarrow \mathrm{v }=\frac{\mathrm{h}}{\mathrm{m\lambda}}=\frac{6.626 \times 10^{-34}}{9.1 \times 10^{-31} \times 4000 \times 10^{10}}=1820 \mathrm{~ms}^{-1} \)

\(v=10^{14} \mathrm{~Hz} ; \mathrm{E}=19.86 \mathrm{~J} \)
\(E=nhv \Rightarrow n=\frac{E}{hv}\)
\(=\frac{19.86}{6.626 \times 10^{-34} \times 10^{14}}=2.99 \times 10^{20} \)
\(\mathrm{n} \simeq 3 \times 10^{20} \)

λ = 5500 x 10^-10 m; P =150 W; Efficiency =12%
\({ E }=\cfrac { hc }{ { \lambda } } =\cfrac { 6.626\times { 10 }^{ -34 }\times 3 \times { 10 }^{ 8 } }{ 5500\times { 10 }^{ -10 } }=3.614 \times 10^{-19} J \)
\(n=\frac{E}{hv}=\cfrac { 150 }{ 3.614\times { 10 }^{ -19 } } =4.15 \times 10^{20}\times\frac{12}{100}\)
n = 4.98 x 10¹⁹s^-1

(i) For a given frequency of incident light the number of photoelectrons emitted is directly proportional to the intensity of the incident light. The saturation current is also directly proportional to the intensity of incident light.
(ii) Maximum kinetic energy of the photo electrons is independent of intensity 0 the incident light.
(iii) Maximum kinetic energy of the photo electrons from a given metal is directly proportional to the frequency of incident light.
(iv) For a given surface, the emission of photoelectrons takes place only if the frequency of incident light is greater than a certain minimum frequency called the threshold frequency.
(v) There is no time lag between incidence of light and ejection of photo electrons.

(i) Davisson - Germer experiment
(ii) Diffraction phenomenon