If the potential difference applied to an X-ray tube is doubled while keeping the separation between the filaments and the target as same, what will happen to the cutoff wavelength?

CONCEPT:

• The instrument which is used for the production of X-ray beams using fast cathode rays is known as X-ray tube.

• For an X-ray tube relation between stopping potential and minimum energy of X-ray beam is can be given as
• \({E_{minimum}} = e{V_0} = \frac{{hc}}{{{\lambda _{cut - off}}}}\)

Where E_minimum = The energy of X-ray beam, V₀ = Stopping potential, h = Plank’s constant, c = Speed of light and λ_cut-off = Cutoff Wavelength

CALCULATION:

Given that

Stopping potential is increases twice hence V₂ = 2V₀ where V₁ = V₀,

∴ Cut off potential when stopping potential is V_o

\({\lambda _{cut - off}} = \frac{{hc}}{{e{V_1}}} = \frac{{hc}}{{e{V_o}}}\)

Cut off potential when stopping potential is doubled, then

Hence

\(\lambda {'_{cut - off\;}} = \frac{{hc}}{{e{V_2}}} =\frac{{hc}}{{e\left( {2{V_0}} \right)}}\; = \frac{{{\lambda _{cut - off}}}}{2}\)