An α-particle and a proton are accelerated from rest by a potential difference of 100V. After this, their de Broglie wavelengths are λα and λp, respectively. The ratio λp/λα, to the nearest integer, is:
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Solution
Kinetic energy = K.E = qV De Broglie wavelength = h√2mK.E = h√2mqV Therefore ratio λpλα = √mαqα√mpqp Alpha particle has 4 times the mass and twice the charge of a proton . λpλα = √8≈ 3