The cut-off wavelength of the characteristic x - rays is
\({ \lambda }_{ ° }\frac { 12400 }{ V } \mathring { A } =\frac { 12400 }{ 20000 } \mathring { A }\)
= 0.62 \(\mathring { A } \)
The corresponding frequency is
\({ \upsilon }_{ o }=\frac { c }{ { \lambda }_{ o } } =\frac { 3\times 10^{ 8 } }{ 0.62\times 10^{ -10 } }\) = 4.84 x 10¹⁸ Hz

\(\lambda_{1}=3310 Å=3310 \times 10^{-10} \mathrm{~m} ; \mathrm{E}_{1}=3 \times 10^{-19} \mathrm{~J} \)
\(\lambda_{2}=5000 Å=5000 \times 10^{-10} \mathrm{~m} ; \mathrm{E}_{2}=0.972 \times 10^{-19} \mathrm{~J} \)
\(\mathrm{E}=\mathrm{E}_{1}-\mathrm{E}_{2}=2.028 \times 10^{-19} J\)
\(\mathrm{hc}\left(\frac{1}{\lambda_{1}}-\frac{1}{\lambda_{2}}\right)=\mathrm{E}_{1}-\mathrm{E}_{2} \)
\(\frac{\mathrm{h} \times 3 \times 10^{8}}{10^{-10}}\left(\frac{1}{3310}-\frac{1}{5000}\right)=2.028 \times 10^{-19} \)
\(\mathrm{~h}=\frac{2.028 \times 10^{-19} \times 10^{-10} \times 3310 \times 5000}{3 \times 10^{8} \times 1690}=6.62 \times 10^{-34} \mathrm{Js} \)
\(\phi_{0} =\frac{\mathrm{hc}}{\lambda}-\mathrm{E}=\frac{6.62 \times 10^{-34} \times 3 \times 10^{8}}{3310 \times 10^{-10}}-3 \times 10^{-19} \)
\(=(6-3) \times 10^{-19}=3 \times 10^{-19} \mathrm{~J} \)
\(\phi_{0} =3 \times 10^{-19} \mathrm{~J} \)
Threshold Wavelength,
\(\lambda_{0}=\frac{\mathrm{hc}}{\phi_{0}}=\frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{3 \times 10^{-19}}=6.62 \times 10^{-7} \mathrm{~m} \)
\(\lambda_{0}=6620 \stackrel {o}{A}\)

(i) λ_v= 413 nm = 413 x 10^-9 m
\({ E }=\cfrac { hc }{ { \lambda } } =\cfrac { 6.626\times { 10 }^{ -34 }\times 3 \times { 10 }^{ 8 } }{ 413\times { 10 }^{ -9 } m} \)
E = 3.00 eV 
(ii) λ = 0.1 nm = 0.1 x 10^-9 m
\({ E }=\cfrac { hc }{ { \lambda } } =\cfrac { 6.626\times { 10 }^{ -34 }\times 3 \times { 10 }^{ 8 } }{ 0.1\times { 10 }^{ -9 } }=19.878\times10^{-16} \) 
E = \(\cfrac { 19.878\times { 10 }^{ -16 } }{ 1.6\times { 10 }^{ -19 } } =12.4237 \ = eV \ 12424 \ eV\) 
E = 12424 eV.
(iii) λ = 10 m
\({ E }_{ r }=\cfrac { hc }{ { \lambda }_{ r } } =\cfrac { 19.878\times { 10 }^{ -27 } }{ 10 } \) 
= \({ E }=\cfrac { hc }{ { \lambda } } =\cfrac { 6.626\times { 10 }^{ -34 }\times 3 \times { 10 }^{ 8 } }{ 10\times 1.6\times{ 10 }^{ -9 } } =1.2424\times { 10 }^{ -7 }\)
E_r = 1.24 x 10^-7 eV.

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 \(\mathring { A }\)

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

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

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

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

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

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