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### IIT-JEE 2010 Paper 1 Offline

Numerical

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 _____________.

Correct Answer is 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

$${{{\lambda _p}} \over {{\lambda _\alpha }}} = \sqrt {{{2{m_\alpha }{q_\alpha }V} \over {2{m_p}{q_p}V}}} = \sqrt {{{2\,.\,4u\,.\,2e\,.\,100} \over {2\,.\,1u\,.\,1e\,.\,100}}}$$

$$= \sqrt 8 = 2.8 \approx 3$$

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