JEE Advance - Physics (2010 - Paper 1 Offline - No. 26)
An $$\alpha$$-particle and a proton are accelerated from the rest by a potential difference of 100 V. After this, their de Broglie wavelengths are $$\lambda$$$$\alpha$$ and $$\lambda$$p, respectively. The ratio $${{{\lambda _p}} \over {{\lambda _\alpha }}}$$, to the nearest integer, is _____________.
Answer
3
Explanation
The de Broglie wavelength of a particle with momentum p is given by
$$\lambda$$ = h/p.
The momentum and kinetic energy of a particle of mass m are related by
$$p = \sqrt {2mK} $$.
The kinetic energy of a charge q, accelerated through potential V, is given by K = qV. Thus,
$$\lambda = h/\sqrt {2mK} = h/\sqrt {2mqV} $$,
which gives
$${{{\lambda _p}} \over {{\lambda _\alpha }}} = \sqrt {{{2{m_\alpha }{q_\alpha }V} \over {2{m_p}{q_p}V}}} = \sqrt {{{2\,.\,4u\,.\,2e\,.\,100} \over {2\,.\,1u\,.\,1e\,.\,100}}} $$
$$ = \sqrt 8 = 2.8 \approx 3$$
Comments (0)
