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1

### JEE Main 2021 (Online) 27th August Evening Shift

Numerical
The number of photons emitted by a monochromatic (single frequency) infrared range finder of power 1 mW and wavelength of 1000 nm, in 0.1 second is x $$\times$$ 1013. The value of x is _____________. (Nearest integer)

(h = 6.63 $$\times$$ 10$$-$$34 Js, c = 3.00 $$\times$$ 108 ms$$-$$1)

## Explanation

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)
2

### JEE Main 2021 (Online) 27th August Morning Shift

Numerical
The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is equal to $${{{h^2}} \over {xma_0^2}}$$. The value of 10x is ___________. (a0 is radius of Bohr's orbit) (Nearest integer) [Given : $$\pi$$ = 3.14]

## Explanation

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

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

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

$$\Rightarrow$$ x = 315.507

$$\Rightarrow$$ 10x = 3155 (nearest integer)
3

### JEE Main 2021 (Online) 26th August Evening Shift

Numerical
A metal surface is exposed to 500 nm radiation. The threshold frequency of the metal for photoelectric current is 4.3 $$\times$$ 1014 Hz. The velocity of ejected electron is ____________ $$\times$$ 105 ms$$-$$1 (Nearest integer)

[Use : h = 6.63 $$\times$$ 10$$-$$34 Js, me = 9.0 $$\times$$ 10$$-$$31 kg]

## Explanation

$$\upsilon$$ : 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.
4

### JEE Main 2021 (Online) 25th July Evening Shift

Numerical
An accelerated electron has a speed of 5 $$\times$$ 106 ms$$-$$1 with an uncertainty of 0.02%. The uncertainty in finding its location while in motion is x $$\times$$ 10$$-$$9 m. The value of x is ____________. (Nearest integer)

[Use mass of electron = 9.1 $$\times$$ 10$$-$$31 kg, h =6.63 $$\times$$ 10$$-$$34 Js, $$\pi$$ = 3.14]

## Explanation

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

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

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

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

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

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