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What is the value of de-broglie wavelength of electron accelerated at 400V?
- a)1.00 nm
- b)0.8 nm
- c)0.06 nm
- d)none of these
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
What is the value of de-broglie wavelength of electron accelerated at ...
Momentum and kinetic energy are related by:
By de-broglie relation:
λ = h/p = h/
KE of electron = q x V = 1.6 x 10-19 x 400 = 6.4 x 10-17 J
m = 9.1 x 10-31Kg
therefore, wavelength = 0.06nm
By de-broglie relation:
λ = h/p = h/
KE of electron = q x V = 1.6 x 10-19 x 400 = 6.4 x 10-17 J
m = 9.1 x 10-31Kg
therefore, wavelength = 0.06nm
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What is the value of de-broglie wavelength of electron accelerated at ...
Calculation of de Broglie wavelength of electron accelerated at 400V
- Given:
Voltage (V) = 400V
Charge of electron (e) = 1.6 x 10^-19 C
Mass of electron (m) = 9.11 x 10^-31 kg
Planck's constant (h) = 6.626 x 10^-34 J s
- Formula:
The de Broglie wavelength of an electron can be calculated using the formula:
λ = h / √(2m eV)
- Calculation:
Substitute the given values into the formula:
λ = (6.626 x 10^-34) / √(2 * 9.11 x 10^-31 * 1.6 x 10^-19 * 400)
λ = 6.626 x 10^-34 / √(2 * 9.11 x 10^-31 * 1.6 x 10^-19 * 400)
λ = 6.626 x 10^-34 / √(2 * 9.11 x 10^-31 * 6.4 x 10^-14)
λ = 6.626 x 10^-34 / √(1.168 x 10^-29)
λ = 6.626 x 10^-34 / 1.08 x 10^-14
λ = 6.13 x 10^-20 m
λ = 0.06 nm
Therefore, the de Broglie wavelength of an electron accelerated at 400V is 0.06 nm. Hence, option (c) is correct.
- Given:
Voltage (V) = 400V
Charge of electron (e) = 1.6 x 10^-19 C
Mass of electron (m) = 9.11 x 10^-31 kg
Planck's constant (h) = 6.626 x 10^-34 J s
- Formula:
The de Broglie wavelength of an electron can be calculated using the formula:
λ = h / √(2m eV)
- Calculation:
Substitute the given values into the formula:
λ = (6.626 x 10^-34) / √(2 * 9.11 x 10^-31 * 1.6 x 10^-19 * 400)
λ = 6.626 x 10^-34 / √(2 * 9.11 x 10^-31 * 1.6 x 10^-19 * 400)
λ = 6.626 x 10^-34 / √(2 * 9.11 x 10^-31 * 6.4 x 10^-14)
λ = 6.626 x 10^-34 / √(1.168 x 10^-29)
λ = 6.626 x 10^-34 / 1.08 x 10^-14
λ = 6.13 x 10^-20 m
λ = 0.06 nm
Therefore, the de Broglie wavelength of an electron accelerated at 400V is 0.06 nm. Hence, option (c) is correct.
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