Question

1.) What transition in He+ has the same frequency (disregarding mass diffrences) as the 2p -> 1s Transition in H?

2.) The workfunction for metallic caesium is 2.14 eV. Calculate the kinetic energy and the speed of the electrons ejected by light of wavelength (a) 750 nm, (b) 250 nm.

Answer 1

1. He⁺ has only one electron so it's like a hydrogen. ransitions in both He⁺ and H can be described by Rydberg’s formula: 1/λ = RZ² (1/n²₁ − 1/n²₂ )

where λ is the wavelength of the emitted photon, R is Rydberg’s constant, Z is the atomic number (1 for H, 2 for He), and n₁ and n₂ are the principal quantum numbers for the initial and final states, respectively.

in order to be allowed, transitions must satisfy the selection rule for the orbital angular momentum quantum number: ∆l = ±1. Thus, the allowed transitions in He+ with the same energy as the 2p→1s transition in H are: 4p→2s, 4s→2p, and 4d→2p.

2.

Kinetic energy = hf - W
h = 6.626068 × 10^-34 J*s = 4.13566733(10)×10−15 eV*s

f = frequency of light (meters)

W = work function = 2.14 eV

1 joule = 6.24150974 × 10¹⁸ eV

2.14 eV = (2.14/6.24150974 × 10¹⁸ ) J = 3.429 x 10^-19 J

c = f x λ, λ = wavelength in m; c = 3.00 x 10⁸ m/s; f = frequency in Hz or s^-1

a) λ= 750 nm = 7.50 x 10^-7 m

f = c/λ = (3.00 x 10⁸)/(7.50 x 10^-7) = 4.00 x 10¹⁴ s^-1

Kinetic energy = (6.626 × 10^-34 )(4.00 x 10¹⁴ ) - 3.429 x 10^-19 J

Kinetic energy = -7.786 x 10^-20 J This wavelength of light will not eject an electron.

b) λ = 250 nm = 2.50 x 10^-7 m

f = c/λ = (3.00 x 10⁸)/(2.50 x 10^-7) = 1.2 x 10¹⁵ s^-1

Kinetic energy = (6.626 × 10^-34)(1.2 x 10¹⁵ s^-1) - 3.429 x 10^-19 J

Kinetic energy = 4.522 x 10^-19 J This wavelength of light will eject an electron.

electron mass = 9.10938188 × 10^-31 kg

Kinetic energy = (0.5)mv²
v² = 2Ek/m = 2(4.522 x 10^-19)/9.10938188 × 10^-31 kg (Ek = kinetic energy)
v² = 9.923 x 10¹¹ m²/s²
v = 9.96 x 10⁵ m/s is the speed of the electron ejected.