SATHEE: Atomic Structure Question 171

Atomic Structure Question 171

Question: An electron has kinetic energy $2.8 \times 10^{- 23} J$ . de-Broglie wavelength will be nearly $(m_{e} = 9.1 \times 10^{- 31} k g)$

Options:

A) $9.28 \times 10^{- 4}, m$

B) $9.28 \times 10^{- 7}, m$

C) $9.28 \times 10^{- 8}, m$

D) $9.28 \times 10^{- 10}, m$

Show Answer

Answer:

Correct Answer: C

Solution:

Formula for de-Broglie wavelength is $λ = \frac{h}{p}$ or $λ = \frac{h}{m v} \Rightarrow e V = \frac{1}{2} m v^{2}$ or $ν = \sqrt{\frac{2 e V}{m}}$

$λ = \frac{h}{\sqrt{2 m e V}}$

$= \frac{6.62 \times 10^{- 34}}{\sqrt{2 \times 9.1 \times 10^{- 31} \times 2.8 \times 10^{- 23}}}$

$λ = 9.28 \times 10^{- 8} m e t e r$ .

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Atomic Structure Question 171

Question: An electron has kinetic energy 2.8×10−23J . de-Broglie wavelength will be nearly (me=9.1×10−31kg)

Options:

Answer:

Solution:

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Question: An electron has kinetic energy $2.8 \times 10^{- 23} J$ . de-Broglie wavelength will be nearly $(m_{e} = 9.1 \times 10^{- 31} k g)$