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Question
An electron is accelerated through a potential of 120 V. Find its de Broglie wavelength.
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Solution
Given: V = 120 V
To find: de Broglie wavelength of the electron
Formula: λ (in nm) = `1.228/sqrt"V"`
Calculation:
From formula,
`lambda = 1.228/sqrt(120)`
= antilog {log (1.228) – 0.5 × log (120)}
= antilog {0.0892 – 0.5 × 2.0792}
= antilog {`overline1`.0496} = 0.1121 nm
The de Broglie wavelength of the electron is 0.1121 nm.
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