No. of different wavelengths = $${{n(n - 1)} \over 2}$$

$$ = {{6 \times (6 - 1)} \over 2} = {{6 \times 5} \over 2} = 15$$

For photon $${\lambda _1} = {h \over P} = {{6.6 \times {{10}^{ - 34}}} \over {{{10}^{ - 27}}}}$$

For particle $${\lambda _2} = {h \over {mv}} = {{6.6 \times {{10}^{ - 34}}} \over {9.1 \times {{10}^{ - 31}} \times {{10}^6}}}$$

$$\therefore$$ $${{{\lambda _1}} \over {{\lambda _2}}} = 910$$

$$I = {e \over T} = {{e\omega } \over {2\pi }} = {{eV} \over {2\pi r}}$$

$$I = {{1.6 \times {{10}^{ - 19}} \times 2.2 \times {{10}^6} \times 7} \over {2 \times 22 \times 0.5 \times {{10}^{ - 10}}}}$$

= 1.12 mA

= 112 $$\times$$ 10^$$-$$2 mA

We know, $$A = {A_0}{e^{ - \lambda t}}$$

For half life

$${{{A_0}} \over 2} = {e^{ - \lambda {t_{1/2}}}}$$

$$ \Rightarrow $$ $${\lambda {t_{1/2}}}$$ = ln 2 .....(1)

And when radioactive element becomes $${\left( {{1 \over 8}} \right)^{th}}$$ of its initial value in 30 years

$${{{A_0}} \over 8} = {A_0}{e^{ - \lambda \times 30}} \Rightarrow \lambda \times 30 = \ln 8$$

$$ \Rightarrow $$ 30$$\lambda = 3\ln 2$$

$$ \Rightarrow $$ $$\lambda = {{3\ln 2} \over {30}}$$ .....(2)

Putting value of $$\lambda $$ in (1), we get

$${{3\ln 2} \over {30}} \times {t_{1/2}}$$ = ln 2

$$ \Rightarrow $$ $${t_{1/2}}$$ = 10 years