An electron is accelerated through a potential difference of 10,000 V. Its de Broglie wavelength is, (nearly) () [2019]

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Q.
An electron is accelerated through a potential difference of 10,000 V. Its de Broglie wavelength is, (nearly) ( $m_{e} = 9 \times 10^{- 31} kg$ ) [2019]

1 12.2 nm

2 $12.2 \times 10^{- 13}$ m

3 $12.2 \times 10^{- 12}$ m

4 $12.2 \times 10^{- 14}$ m

Ans.
(3)

de Broglie wavelength of electron,

$λ_{e} = \frac{12.27 Å}{\sqrt{V (in V)}}$

Here, $V = 10000$ $V$

$∴$ $λ_{e} = \frac{12.27}{\sqrt{10000}} \times 10^{- 10} m = 12.27 \times 10^{- 12} m$

Solution

Q.
An electron is accelerated through a potential difference of 10,000 V. Its de Broglie wavelength is, (nearly) ( $m_{e} = 9 \times 10^{- 31} kg$ ) [2019]

1 12.2 nm

2 $12.2 \times 10^{- 13}$ m

3 $12.2 \times 10^{- 12}$ m

4 $12.2 \times 10^{- 14}$ m

Ans.
(3)

de Broglie wavelength of electron,

$λ_{e} = \frac{12.27 Å}{\sqrt{V (in V)}}$

Here, $V = 10000$ $V$

$∴$ $λ_{e} = \frac{12.27}{\sqrt{10000}} \times 10^{- 10} m = 12.27 \times 10^{- 12} m$

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Solution

Q. An electron is accelerated through a potential difference of 10,000 V. Its de Broglie wavelength is, (nearly) (me=9×10-31kg) [2019]

1 12.2 nm

2 12.2×10-13m

3 12.2×10-12m

4 12.2×10-14m

Ans. (3) de Broglie wavelength of electron, λe=12.27 ÅV (in V) Here, V=10000 V ∴ λe=12.2710000×10-10m=12.27×10-12m

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Q.
An electron is accelerated through a potential difference of 10,000 V. Its de Broglie wavelength is, (nearly) ( $m_{e} = 9 \times 10^{- 31} kg$ ) [2019]

2 $12.2 \times 10^{- 13}$ m

3 $12.2 \times 10^{- 12}$ m

4 $12.2 \times 10^{- 14}$ m

Ans.
(3)

de Broglie wavelength of electron,

$λ_{e} = \frac{12.27 Å}{\sqrt{V (in V)}}$

Here, $V = 10000$ $V$

$∴$ $λ_{e} = \frac{12.27}{\sqrt{10000}} \times 10^{- 10} m = 12.27 \times 10^{- 12} m$