NEET-XII-Physics
42: Photoelectric Effect and Wave Particle Duality
- #12-athe same speed (b) the same momentum (c) the same energy?Ans : It is given that the speed of an electron and proton are equal.
It is clear from the above equation that
`` \lambda \propto \frac{1}{m}``
As mass of a proton, mp > mass of an electron, me, the proton will have a smaller wavelength compared to the electron. (b) `` \lambda =\frac{h}{p}`` `` \left(\because p=mv\right)``
So, when the proton and the electron have same momentum, they will have the same wavelength. (c) de-Broglie wavelength `` \left(\lambda \right)`` is also given by
`` \lambda =\frac{h}{\sqrt{2mE}}``,
where E = energy of the particle.
Let the energy of the proton and electron be E.
Wavelength of the proton,
`` {\lambda }_{p}=\frac{h}{\sqrt{2{m}_{\,\mathrm{\,p\,}}E}}...\left(1\right)``
Wavelength of the electron,
`` {\lambda }_{e}=\frac{h}{\sqrt{2{m}_{\,\mathrm{\,e\,}}E}}...\left(2\right)``
Dividing (2) by (1), we get:
`` \frac{{\lambda }_{e}}{{\lambda }_{p}}=\frac{\sqrt{{m}_{e}}}{\sqrt{{m}_{p}}}``
`` \Rightarrow \frac{{\lambda }_{e}}{{\lambda }_{p}}<1``
`` \Rightarrow {\lambda }_{e}<{\lambda }_{p}``
It is clear that the proton will have smaller wavelength compared to the electron.
Page No 363:
- #12-bthe same momentumAns : `` \lambda =\frac{h}{p}`` `` \left(\because p=mv\right)``
So, when the proton and the electron have same momentum, they will have the same wavelength.
- #12-cthe same energy?Ans : de-Broglie wavelength `` \left(\lambda \right)`` is also given by
`` \lambda =\frac{h}{\sqrt{2mE}}``,
where E = energy of the particle.
Let the energy of the proton and electron be E.
Wavelength of the proton,
`` {\lambda }_{p}=\frac{h}{\sqrt{2{m}_{\,\mathrm{\,p\,}}E}}...\left(1\right)``
Wavelength of the electron,
`` {\lambda }_{e}=\frac{h}{\sqrt{2{m}_{\,\mathrm{\,e\,}}E}}...\left(2\right)``
Dividing (2) by (1), we get:
`` \frac{{\lambda }_{e}}{{\lambda }_{p}}=\frac{\sqrt{{m}_{e}}}{\sqrt{{m}_{p}}}``
`` \Rightarrow \frac{{\lambda }_{e}}{{\lambda }_{p}}<1``
`` \Rightarrow {\lambda }_{e}<{\lambda }_{p}``
It is clear that the proton will have smaller wavelength compared to the electron.
Page No 363: