NEET-XII-Physics
42: Photoelectric Effect and Wave Particle Duality
- #12Consider the de-Broglie wavelength of an electron and a proton. Which wavelength is smaller if the two particles have (a) the same speed (b) the same momentum (c) the same energy? (a) the same speed (b) the same momentum (c) the same energy?Ans : de-Broglie wavelength,
`` \lambda =\frac{h}{mv}``,
where h = Planck's constant
`` m`` = mass of the particle
v = velocity of the particle (a) 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: (a) 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-athe same speedAns : 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.
- #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: