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Answer 1

QUESTION:

Calculate the de Broglie wavelength for

(i) A 6.626 kg iron ball moving at 10 m/s.

(ii) An electron moving at 72.73 m/s.

ANSWER:

The de Broglie wavelength

• (i) $\sf \lambda = {10}^{ - 35}\ m$

• (ii) $\sf \lambda = {10}^{ -5}\ m$

TO FIND:

• The de Broglie wavelength for the given cases.

GIVEN:

• (i) A 6.626 kg iron ball moving at 10 m/s.

• (ii) An electron moving at 72.73 m/s.

EXPLANATION:

$\boxed{\bold{De - Broglie Wavelength = \dfrac{h}{mv}}}$

(i) A 6.626 kg iron ball moving at 10 m/s.

$\sf h = 6.626 \times {10}^{ - 34} \ Js$

$\sf m = 6.626 \ kg$

$\sf v = 10\ m/s$

$\sf \lambda = \dfrac{6.626 \times {10}^{ - 34} }{6.626 \times 10}$

$\sf \lambda = \dfrac{ {10}^{ - 34} }{ 10}$

$\sf \lambda = {10}^{ - 35}\ m$

(ii) An electron moving at 72.73 m/s.

$\sf h = 6.626 \times {10}^{ - 34} \ Js$

$\sf Mass\ of\ electron = 9.11 \times 10^{-31} \ kg$

$\sf v = 72.73 \ m/s$

$\sf \lambda = \dfrac{6.626 \times {10}^{ - 34} }{9.11 \times 10^{-31} \times 72.73 }$

$\sf 9.11 \times 72.73 \approx 662.6$

$\sf \lambda = \dfrac{6.626 \times {10}^{ - 34} }{662.6 \times 10^{-31} }$

$\sf \lambda = \dfrac{ {10}^{ - 34} }{100 \times 10^{-31} }$

$\sf \lambda = \dfrac{ {10}^{ - 34} }{ 10^{-29} }$

$\sf \lambda = {10}^{ - 5} \ m$