JEE MAIN - Chemistry (2021 - 16th March Morning Shift - No. 21)
When light of wavelength 248 nm falls on a metal of threshold energy 3.0 eV, the de-Broglie wavelength of emitted electrons is _______$$\mathop A\limits^o $$. (Round off to the Nearest Integer).
[ Use : $$\sqrt 3 $$ = 1.73, h = 6.63 $$\times$$ 10$$-$$34 Js
me = 9.1 $$\times$$ 10$$-$$31 kg; c = 3.0 $$\times$$ 108 ms$$-$$1; 1eV = 1.6 $$\times$$ 10$$-$$19 J]
[ Use : $$\sqrt 3 $$ = 1.73, h = 6.63 $$\times$$ 10$$-$$34 Js
me = 9.1 $$\times$$ 10$$-$$31 kg; c = 3.0 $$\times$$ 108 ms$$-$$1; 1eV = 1.6 $$\times$$ 10$$-$$19 J]
Answer
9
Explanation
$$\lambda$$ = 248 $$\times$$ 10$$-$$9 m
w0 = 3 $$\times$$ 1.6 $$\times$$ 10$$-$$19 J
E = w0 + K.E.
$${{hc} \over \lambda } = {W_0} + K.E.$$
$$K.E. = {{6.626 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {248 \times {{10}^{ - 19}}}} - 3 \times 1.6 \times {10^{ - 19}}$$
= 3.2 $$\times$$ 10$$-$$19 J
p = $$\sqrt {2mK.E.} $$
p = $$\sqrt {2 \times 9.1 \times {{10}^{ - 31}} \times 3.2 \times {{10}^{ - 19}}} $$
p = 7.63 $$\times$$ 10$$-$$25
$$ \therefore $$ $$\lambda = {h \over p} = {{6.626 \times {{10}^{ - 34}}} \over {7.63 \times {{10}^{ - 25}}}}$$
$$ \Rightarrow $$ $$\lambda$$ = 8.7 $$\times$$ 10$$-$$10 = 8.7 $$\mathop A\limits^o $$ $$ \approx $$ 9 $$\mathop A\limits^o $$
w0 = 3 $$\times$$ 1.6 $$\times$$ 10$$-$$19 J
E = w0 + K.E.
$${{hc} \over \lambda } = {W_0} + K.E.$$
$$K.E. = {{6.626 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {248 \times {{10}^{ - 19}}}} - 3 \times 1.6 \times {10^{ - 19}}$$
= 3.2 $$\times$$ 10$$-$$19 J
p = $$\sqrt {2mK.E.} $$
p = $$\sqrt {2 \times 9.1 \times {{10}^{ - 31}} \times 3.2 \times {{10}^{ - 19}}} $$
p = 7.63 $$\times$$ 10$$-$$25
$$ \therefore $$ $$\lambda = {h \over p} = {{6.626 \times {{10}^{ - 34}}} \over {7.63 \times {{10}^{ - 25}}}}$$
$$ \Rightarrow $$ $$\lambda$$ = 8.7 $$\times$$ 10$$-$$10 = 8.7 $$\mathop A\limits^o $$ $$ \approx $$ 9 $$\mathop A\limits^o $$
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