Question

2. What should be the energy of an electron so that the associated electron waves have a wavelength of 600 nm? ince the visible region spans between approximately 400 nm and 700 nm, why can the electron wave mentioned in Problem 2 not be seen by the human eye? What kind of device is necessary to detect electron waves? 4. What is the energy of a light quantum (photon) which has a wavelength of 600 nm? Compare the energy with the electron wave energy calculated in Problem 2 and discuss the difference. le uith a velocity of 200 km/h. What is the

The answer of number 2 is E=4.18 *10^-6 ev

And number 4 is E=2.07 ev

Answer 1

$\text{Solution : }\\ Q \ no.2\\ \text{Relation between the energy of particle and its De Broglie wavelength is given by }\\ \lambda = \dfrac{h}{\sqrt{2m\ E}}\\ or\\ E = \dfrac{h^2}{2m \ \lambda^2}\\ \ \\ We\ have :\\ \text{Wavelength} \ \lambda = 600nm= 600\times 10^{-9}m, \\ \ \ \ \ \text{Plank's \ constant }h = 6.626\times 10^{-34}J.s \\ \ \ \ \text{Mass of the electron} \ m = 9.1\times 10^{-31}kg \\ \text{Putting these values in the above equation, we get }\\ Energy;\\ E_{electron} = \dfrac{(6.626\times 10^{-34}J.s)^2}{2\times (9.1\times 10^{-31}kg)\times(600\times 10^{-9}m)^2}\\ E_{electron}= 6.70\times 10^{-25}J\\ E_{electron} = 6.70\times 10^{-23}\times(6.242\times 10^{18}eV)\\ E_{electron} = 4.18\times10^{-6}eV\\ \ \\ Q \ no. 3\\ \text{The electron wave is a matter wave. It is not an electromagnetic wave. Our Eyes is sensitive to only that the range of electromagnetic waves that fall in the visible region. }$

$\text{In davission Germer experiment , electron waves is observed by phenomena of diffraction.}$

$Q \ no. 4\\ \text{Energy of photon is :}\\ E_{photon} = \dfrac{h\ c}{\lambda}\\ E_{photon} = \dfrac{(6.626\times 10^{-34}J.s)\times(3\times10^8m/s)}{600\times 10^{-9}m}\\ E_{photon} = 3.13\times 10^{-19}J\\ E_{photon} = 3.13\times(6.242\times 10^{18}eV)\\ E_{photon} = 2.07\ eV\\ \ \\ E_{photon} > E_{electron}$