Energy emitted in 0.1 sec.

= 0.1 sec. $$\times$$ $${10^{ - 3}}{J \over s}$$

= 10^$$-$$4 J

If 'n' photons of $$\lambda$$ = 1000 nm are emitted, then 10^$$-$$4 = n $$\times$$ $${{hc} \over \lambda }$$

$$ \Rightarrow {10^{ - 4}} = {{n \times 6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {1000 \times {{10}^{ - 9}}}}$$

$$\Rightarrow$$ n = 5.02 $$\times$$ 10¹⁴ = 50.2 $$\times$$ 10¹³

$$\Rightarrow$$ 50 (nearest integer)

$$mvr = {{nh} \over {2\pi }}$$

$$K.E. = {{{n^2}{h^2}} \over {8{\pi ^2}m{r^2}}} = {{4{h^2}} \over {8{\pi ^2}m{{(4{a_0})}^2}}}$$

$$ = \left( {{4 \over {8{\pi ^2} \times 16}}} \right){{{h^2}} \over {ma_0^2}}$$

$$\Rightarrow$$ x = 315.507

$$\Rightarrow$$ 10x = 3155 (nearest integer)

$$\upsilon $$ : speed of electron having max. K.E.

$$\Rightarrow$$ from Einstein equation : E = $$\phi$$ + K.E._max

$$ \Rightarrow {{hc} \over \lambda } = h{\upsilon _0} + {1 \over 2}m{v^2}$$

$$ \Rightarrow {{6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {500 \times {{10}^{ - 9}}}} = 6.63 \times {10^{ - 34}} \times 4.3 \times {10^{14}} + {1 \over 2}m{v^2}$$

$$ \Rightarrow {{6.63 \times 30 \times {{10}^{ - 20}}} \over 5} = 6.63 \times 4.3 \times {10^{ - 20}} + {1 \over 2}m{v^2}$$

$$ \Rightarrow 11.271 \times {10^{ - 20}}J = {1 \over 2} \times 9 \times {10^{ - 31}} \times {\upsilon ^2}$$

$$\Rightarrow$$ $$\upsilon $$ = 5 $$\times$$ 10⁵ m/sec.

$$\Delta v = {{0.02} \over {100}} \times 5 \times {10^6} = {10^3}$$ m/s

$$\Delta x.\Delta v = {h \over {4\pi m}}$$

$$ \Rightarrow $$ $$x \times {10^{ - 9}} \times {10^3} = {{6.63 \times {{10}^{ - 34}}} \over {4 \times 3.14 \times 9.1 \times {{10}^{ - 31}}}}$$

$$ \Rightarrow $$ $$x \times {10^{ - 9}} \times {10^3} = 0.058 \times {10^{ - 3}}$$

$$ \Rightarrow $$ $$x = {{0.058 \times {{10}^{ - 6}}} \over {{{10}^{ - 9}}}} = 58$$