Photoelectric Effect, What are the threshold wavelength for sodium? Note that wo

Question

Photoelectric Effect, What are the threshold wavelength for sodium? Note that work function for sodium is 3.78 times 10^-19 J and Planck constant is h = 6.626 times 10^-34 J s. What s the stopping potential for sodium using 400-nm violet light? A photoelectric experiment uses 265-nm light and a silver target. Note that work function for silver is 7.43 times 10^-19 J. Find the maximum kinetic energy of the photoelectrons and the stopping potential. Wave-Particle Duality, Find the speed of an electron with de Broglie wavelength 1.0 nm. Note that mass of electron is 9.11 times 10^-31 kg. An electron microscope calls for electrons with de Broglie wavelength 0.25 nm. What's the electrons' speed? Rydberg Equation, Use the Rydberg equation to explore the wavelength pattern for the first two Balmer series (final state is n = 2) and for the first two Paschen series (final state is n = 3). Note that R_H = 0.01097 nm^-1. The Bohr Atom, A hydrogen atom at rest makes a transition from the n = 4 state to n = 2. Find the wavelength of the emitted photon An electron in a hydrogen atom orbits with radius of 4

Explanation / Answer

Threshold wavelength hc/(lambda) = workfunction = 6.626*10^-34 *3* 10^8 / 3.78*10^-19 = LAMBDA

lambda = 5.25*10^-7 m = 525 nm

Stopping potentiAL IS (6.626*10^-34 *3* 10^8 / 400*10^-9) - ( 3.78*10^-19) = (1.6 *10^-19)*V

So V = 0.743V

For silver (6.626*10^-34 *3* 10^8 / 265*10^-9) - ( 7.43*10^-19) = (1.6 *10^-19)*V

V = 0.04 volts

Now max v = sqrt(2 *e*V/m)

and K>E max = e*V = 1.6*10^-19 * 0.04J

debroglie wavelength = h/mv = (6.634*10^-34 / (9.1*10^-31 * v)

v= 7.3 * 10^5 m/s

If wavelength is 0.25nm then velocity willbe4 times = 22.2* 10^5 m/s

Balmer series n =2 From 3 to 2 it willbe 0.01097 * (1/2^2 -1/3^2) = 5/36 * 0.01097 nm^-1. So lambda will be 7.2 /0.01097 nm

For n=4 to 2 it willbe 0.01097 * 3/16 nm^-1 lambda will be 16/(3*0.01097) nm

For paschen series n =3

So for 4 to 3 lambda will be 144 /(7 * 0.01097) nm

and from 5 to 3 we 225/(16*0.01097) nm

Problem 4 : wavelength = 16/(3 * 0.01097) nm

Electrons in hydrogen rotates in orbits of radii propotion to square of prinicipial quantum number n.

So Electron has jumped from n =2 to n =3

we have 1/lambda = 0.01097 * (5/36)

Energy = hc/lambda = 6.634 * 10^-34 * 3 8 10^8 * 0.1097 * (5/36) * 10^9 J

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