Question

Plz do number 4, number 2 answer is 4.18*10^-6 ev

2. What should be the energy of an electron so that the associated electron waves have a wavelength of 600 nm? 3. Since 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. with a valncitv of 200 km/h. What is the

Answer 1

Energy of light quantum: E = hc/lambda Here, h is Plank's constant and c is speed of light hc = 1240 eV n Substitute the numerical values E = 1240 eV.n/600 nm = 2.07 eV In part (2), From debroglie equation, the wavelength equation for electron, lambda = h/squareroot 2E Rearrange for E E = h^2/2m lambda^2 Substitute the numerical values E = (6.624 times 10^-34 J.s)^2/2(9.1 times 10^-31 kg) (600 times 10^-9 m)^2 = 6.69679 times 10^-25 J (1 eV/1.6 times 10^-19 J) = 4.18 times 10^6 eV From the above calculations, it is clear that the energy of photon is much greater than the energy of electron. Mass of the electron is greater than photon (consider only rest mass) and mass inversely proportional to energy, so electron has less energy for same wavelength.