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Find the de Broglie wavelength of Earth. Mass of Earth is 6\times 10^{24} \ kg. Mean orbital radius of Earth around Sun is 150\times 10^{6} \ km -

Option: 1

3.7 \ m


Option: 2

3.7\times10^{-63} \ m


Option: 3

3.7\times10^{63} \ m


Option: 4

3.7\times10^{-63} \ cm


Answers (1)

The Earth revolves around the Sun once a year and its orbit has a mean radius 150 \times 10^{6} \ km .
Hence its average orbital speed is
v=\frac{2 \pi \times 150 \times 10^9}{365 \times 24 \times 3600}=3 \times 10^4 \mathrm{~m} \mathrm{~s}^{-1}
As the Earth has mass 6 \times 10^{24}\ kg, its average momentum is
p = mv = 1.8 \times 10^{29} kg \ m \ s^{-1}
and so its de Broglie wavelength is
\lambda=\frac{\mathrm{h}}{\mathrm{p}}=\frac{6.63 \times 10^{-34}}{1.8 \times 10^{29}}=3.7 \times 10^{-63} \mathrm{~m} \text {, }
If the Sun can be considered stationary.

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Kshitij

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