Correct Answer - Option 1 : 0.1
CONCEPT:
The wavelength of any charged particle due to its motion is called the de-Broglie wavelength.
When a charged particle is accelerated in a potential difference the energy gained by the particle is given by:
Energy (E) = e × V
Where V is the potential difference and q is a charge.
The de-Broglie wavelength of charge particle (λd) is given by:
\({\lambda _b} = \frac{h}{{\sqrt {2m\;E} }}= \frac{h}{{\sqrt {2m\;q\;V} }}\)
Where E is energy, h is Planck constant, m is the mass of the charged particle
EXPLANATION:
De-Broglie wavelength when a charge q is accelerated by a potential difference of V volts is
\({\lambda _b} = \frac{h}{{\sqrt {2qVm} }}\) …i)
For cut off wavelength of X-rays, we have
\(qV = \frac{{hc}}{{{\lambda _m}}}\)
Or \({\lambda _m} = \frac{{hc}}{{qV}}\) …ii)
From Eqs, i) and ii), we get
\(\frac{{{\lambda _b}}}{{{\lambda _m}}} = \frac{{\sqrt {\frac{{qV}}{{2m}}} }}{c}\)
For electron \(\frac{q}{m} = 1.8 \times {10^{11}}\;C/kg\) (given).
Substituting the values the desired ratio is
\(\frac{{{\lambda _b}}}{{{\lambda _m}}} = \frac{{\sqrt {\frac{{1.8\; \times\; {{10}^{11}}\; \times\; 10\; \times\; {{10}^3}}}{2}} }}{{3\; \times\; {{10}^8}}} = 0.1\)
