Question

Part A

An electron undergoes a transition from an initial (n_i) to a final (n_f ) energy state. The energies of the n_i and n_f energy states are −2.420×10⁻¹⁹J and −8.720×10⁻²⁰ J, respectively. Calculate the wavelength of the light in nanometers corresponding to this transition.

You can use the following values for your calculations: Planck′s constant (h) = 6.626×10⁻³⁴ J⋅s, speed of light (c) = 2.998×10⁸ m/s, and 1 m=10⁹ nm Express your answer in nanometers to four significant figures.

Part B

Every electron in an atom is described by a unique set of four quantum numbers: n, ℓ, mℓ, and ms. The principal quantum number, n, identifies the shell in which the electron is found. The angular momentum quantum number, ℓ, indicates the kind of subshell. The magnetic quantum number, mℓ, distinguishes the orbitals within a subshell. The spin quantum number, ms, specifies the electron spin.

Identify which sets of quantum numbers are valid for an electron. Each set is ordered (n,ℓ,mℓ,ms).

Check all that apply.

2,0,0,-1/2

3,3,-2,-1/2

1,0,0,1/2

4,3,1,-1/2

4,3,-4,1/2

3,-2,-2,-1/2

0,2,0,1/2

3,2,-2,1/2

1,1,0,1/2

1,1,0,-1/2

2,1,-1,1/2

3,-2,-1,0

Part C

Identify the sets of quantum numbers that describe all the electrons in a neutral beryllium atom (Be) in the ground state. Each set is ordered (n,ℓ,mℓ,ms).

Drag the appropriate items to their respective bins.

Reset Help 2,0,0,1/2 2,1,0,-1/2 2,1,-1-1/2 1,0,0,-1/2 2,0,0,-1/2 2,1,0,1/2 1,0,0,1/2 2,1,-1,1/2 Electrons in Be Electrons not in Be

Answer 1

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Part A = n; -2420 810'95_mez -8.320x16 we know._AE = be whene .Eç - Ej- | -8-220X10*20 - (- 2.420 410-403] J = 6.626x10";s x 2.999x16mly de 6.626x10^34Js 2.998x13m/s 1-283x10 ºm 1.545x 10193 OD 10 9 = lom so 1.29 3x10-6m = 1-283x10 mx enm - 1283.nm. 10-9m Part B n m s Valid 2 WN 3 o 3 . - Owlo Invalid - Orbital as 36 IS Y 2 -cü - 4f 4 4 3 3 3 -2 o z -4 -2 o 72 ½ % * 46_(but 1-3m3-3.+0+3) L (for l. the values are there) not possible_n= 3d o » i i OK 1 -y +42 3 -2 -1 0 2 LIP IP 22 (for the value are there) v Electrons in Be 2,0,0,* (282 1,0,0 % (18) 2,0,0, (252 1,0,0, Y2 (15) Electron.netin Be Configurationale 2.1.0,-) (29)___152 282DB 21,-1,-% (24,2 2,1,0, +/2 (2/2) 2,1,-1,42 (22) CamScanner