Question

A. A color television tube also generates some X-rays when its electron beam strikes the screen. What is the shortest wavelength of these X-rays in picometers, if a 34.6 kV potential is used to accelerate the electrons? (Note that TVs have shielding to prevent these x rays from exposing viewers.)

B. What is the smallest angle in degrees that L makes with the z -axis for an l = 1 electron?

C. What is the shortest wavelength x-ray radiation (in picometers) that can be generated in an x-ray tube with an applied voltage of 42.5 kV?

D. Calculate the wavelength, in nanometers, of emitted light from hydrogen as the electron's energy state goes from n = 5 to n = 1. Rydberg Constant is 1.097×10⁷ m^-1.

E.An X-ray tube has an applied voltage of 227 kV. What is the most energetic x-ray photon it can produce? Give your answer in keV.

F. How many electrons can be in the n = 5 shell?

G.Calculate the magnitude of the angular momentum, in 10^-34 Js, for an l = 4 electron.

H.If a hydrogen atom has its electron in the n = 6 state, how much energy in eV is needed to ionize it?

H.

Answer 1

Answer:

(A) The shortest wavelength means, the energy of the X-rays is maximum.

That means, _min = hc/E_max

Given that, V = 34.6 kV,

In terms of energy E = 34.6 x 10³ eV

Thus, _min = (6.626 x 10^-34 J.s) (3 x 10⁸ m/s)/(34.6 x 10³) (1.6 x 10^-19 J) = 0.359 x 10^-10 m or 0.359 A⁰

(B) L_z = L cos

or = cos^-1(L_z/L) = cos^-1 { m_l / [l(l+1)]^1/2 }

Given, l = 1 so its m_l values are +1,0 and -1

If we substitute +1 value in the angle, then we get the smallest angle.

Thus, = cos^-1 { +1/ [1(1+1)]^1/2 } = 45.0⁰

(C) Same precedure follows as we did part (A).

_min = hc/E_max

Given, V = 42.5 kV

So, E = 42.5 x 10³ eV

_min = (6.626 x 10^-34 J.s) (3 x 10⁸ m/s)/(42.5 x 10³) (1.6 x 10^-19 J) = 0.292 x 10^-10 m or 0.292 A⁰

(D) For Hydrogen atom,

Energy values are E_n = -13.6 eV/n²

For ground state n = 1,E₁ = -13.6 eV

For n = 5, E₅ = -13.6/25 eV = -0.544 eV

Thus, energy difference E = E₅ - E₁ = -0.544 eV- (-13.6 eV) = 13.056 eV

In terms of Joules, E = 13.056 x 1.6 x 10^-19 J = 20.889 x 10^-19 J

Therefore, the wavelength of the emitted photon is

= hc/E = (6.626 x 10^-34 J.s) (3 x 10⁸ m/s)/(20.889 x 10^-19 J) = 95.16 x 10^-9 m or 95.16 nm