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The wavelength associated with an electron,accelerated through a potential difference of 100V, is of the order of [1996]
  • a)
    1000 Å
  • b)
    100 Å
  • c)
    10.5 Å
  • d)
    1.2 Å
Correct answer is option 'D'. Can you explain this answer?
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The wavelength associated with an electron,accelerated through a poten...
Potential difference = 100 V
K.E. acquired by electron = e (100)
According to de Broglie's concept
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The wavelength associated with an electron,accelerated through a poten...
The wavelength associated with an electron accelerated through a potential difference of 100V can be calculated using the de Broglie wavelength equation:

λ = h / √(2meV)

Where:
λ = wavelength
h = Planck's constant (6.63 x 10^-34 J*s)
me = mass of electron (9.11 x 10^-31 kg)
V = potential difference (100V)

Plugging in the values:

λ = (6.63 x 10^-34 J*s) / √(2 * (9.11 x 10^-31 kg) * (100V))

λ ≈ 1.22 x 10^-9 m

The order of magnitude for this wavelength is 10^-9 m, which is equivalent to 1 nanometer (nm).

Therefore, the answer is:
a) 1000 nm
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The wavelength associated with an electron,accelerated through a potential difference of 100V, is of the order of [1996]a)1000 Åb)100 Åc)10.5 Åd)1.2 ÅCorrect answer is option 'D'. Can you explain this answer?
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