#### 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

will remain unchanged

will be doubled

will be halved

will become four times the original

#### 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.