NEET-XII-Physics
42: Photoelectric Effect and Wave Particle Duality
- #14A proton and an electron are accelerated by the same potential difference. Let λe and λp denote the de Broglie wavelengths of the electron and the proton, respectively.
(a) λe = λp
(b) λe < λp
(c) λe > λp
(d) The relation between λe and λp depends on the accelerating potential difference.digAnsr: cAns : (c) λe > λp
Let me and mp be the masses of electron and proton, respectively.
Let the applied potential difference be V.
Thus, the de-Broglie wavelength of the electron,
`` {\lambda }_{e}=\frac{h}{\sqrt{2{m}_{\,\mathrm{\,e\,}}eV}}...\left(1\right)``
And de-Broglie wavelength of the proton,
`` {\lambda }_{p}=\frac{h}{\sqrt{2{m}_{\,\mathrm{\,p\,}}eV}}...\left(2\right)``
Dividing equation (2) by equation (1), we get:
`` \frac{{\lambda }_{\,\mathrm{\,p\,}}}{{\lambda }_{\,\mathrm{\,e\,}}}=\frac{\sqrt{{m}_{\,\mathrm{\,e\,}}}}{\sqrt{{m}_{\,\mathrm{\,p\,}}}}``
me < mp
`` \therefore `` `` \frac{{\lambda }_{p}}{{\lambda }_{e}}<1``
`` \Rightarrow {\lambda }_{p}<{\lambda }_{e}``
Page No 364: