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Question
If the potential difference applied to the tube is doubled and the separation between the filament and the target is also doubled, the cutoff wavelength
Options
will remain unchanged
will be doubled
will be halved
will become four times the original
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Solution
will be halved
Cut off wavelength is given by `lambda_min = (hc)/(eV)`,
where h = Planck's constant
c = speed of light
e = charge on an electron
V = potential difference applied to the tube
When potential difference (V) applied to the tube is doubled, cutoff wavelength (`lambda'_min`) is given by
`lambda'_min` = `(hc)/(e(2V)`
⇒ `lambda'_min = lambda_min/2`
Cuttoff wavelength does not depend on the separation between the filament and the target.
Thus, cutoff wavelength will be halved if the the potential difference applied to the tube is doubled.
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