NEW
New Website Launch
Experience the best way to solve previous year questions with mock tests (very detailed analysis), bookmark your favourite questions, practice etc...
1

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 _____________.

Your Input ________

Answer

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

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

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

Joint Entrance Examination

JEE Main JEE Advanced WB JEE

Graduate Aptitude Test in Engineering

GATE CSE GATE ECE GATE EE GATE ME GATE CE GATE PI GATE IN

Medical

NEET

CBSE

Class 12