JEE MAIN - Chemistry (2021 - 20th July Evening Shift - No. 18)
The wavelength of electrons accelerated from rest through a potential difference of 40 kV is x $$\times$$ 10$$-$$12 m. The value of x is ___________. (Nearest integer)
Give : Mass of electron = 9.1 $$\times$$ 10$$-$$31 kg
Charge on an electron = 1.6 $$\times$$ 10$$-$$19 C
Planck's constant = 6.63 $$\times$$ 10$$-$$34 Js
Give : Mass of electron = 9.1 $$\times$$ 10$$-$$31 kg
Charge on an electron = 1.6 $$\times$$ 10$$-$$19 C
Planck's constant = 6.63 $$\times$$ 10$$-$$34 Js
Answer
6
Explanation
Wavelength of electron is given by
$$\lambda = {h \over {\sqrt {2mqV} }}$$
Here q = charge on electron, V = potential difference
$$\lambda = {{6.63 \times {{10}^{ - 34}}} \over {\sqrt {2 \times 9.1 \times {{10}^{ - 31}} \times 1.6 \times {{10}^{ - 19}} \times 40 \times {{10}^3}} }}$$
$$ = {{6.63 \times {{10}^{ - 34}}} \over {\sqrt {1164.8 \times {{10}^{ - 47}}} }} = 6.144 \times {10^{ - 12}} \approx 6 \times {10^{ - 12}}$$
x = 6
OR
$$\lambda = {{12.3} \over {\sqrt V }}\mathop A\limits^o $$
$$ = {{12.3} \over {200}} = 6.15 \times {10^{ - 12}}$$ m
$$ \therefore $$ Ans. is 6
$$\lambda = {h \over {\sqrt {2mqV} }}$$
Here q = charge on electron, V = potential difference
$$\lambda = {{6.63 \times {{10}^{ - 34}}} \over {\sqrt {2 \times 9.1 \times {{10}^{ - 31}} \times 1.6 \times {{10}^{ - 19}} \times 40 \times {{10}^3}} }}$$
$$ = {{6.63 \times {{10}^{ - 34}}} \over {\sqrt {1164.8 \times {{10}^{ - 47}}} }} = 6.144 \times {10^{ - 12}} \approx 6 \times {10^{ - 12}}$$
x = 6
OR
$$\lambda = {{12.3} \over {\sqrt V }}\mathop A\limits^o $$
$$ = {{12.3} \over {200}} = 6.15 \times {10^{ - 12}}$$ m
$$ \therefore $$ Ans. is 6
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