1) Calculate the momentum of an electron traveling 46.9% the speed of light. Rep

Question

1) Calculate the momentum of an electron traveling 46.9% the speed of light. Report answer in kg(m/s) to 3 significant digits. Do not put units in answer box.

2)Which quantum number(s) determine the energy of a hydrogen like atom using the Bohr model?

Select one or more:

A. n

B. Rh

C. l

D. m

3)Calculate the change in energy of an electron in a hydrogen atom going from the ground state to the first excited state. Report answer in J to 3 significant digits.

4) Calculate the energy of photon emitted from a hydrogen atomin the n = 5 state relaxing to the n = 3 state. Report answer in J to 3 significant digits.

5) How many orbitals are there in the 6d subshell?

Explanation / Answer

Q1

46.9% of the speed of light is 0.469 * (3 x 108 m/s) = 1.407 x 108 m/s

Momentum is given by p = mv = mev = 9.109 x 10-31 kg ( 1.407 x 108 m/s) = 1.28 x 10-22 kg m s-1

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Q2

The principle qunatum number n determines the energy. Thus the answer is a.

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Q3

Using the Rhydberg equation we get:

1/lambda = -R (nf-2 - ni-2)

= -1.097 x 107 m-1(2-2 - 1-2) = 8227500 m-1

lambda = 1.215 x 10-7 m

The sign indicates

Solving E = hc/lambda for energy gives E = 1.64 x 10-18 J.

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Q4

Applying the same equation for nf = 3 and ni = 5. The sign of E will be negative because energy is released.

1/lambda = -R (nf-2 - ni-2)

= -1.097 x 107 m-1(3-2 - 5-2) = -780089 m-1 = 1 / lambda

Lambda = -1.28 x 10-6 m

Solving E = hc/lamda gives E = -1.55 x 10-19 J. This means emission.

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Q5

For a 6d subshell, we see that n = 6 and that l = 2 for a d orbital.

That means ml = -2, -1, 0, 1, 2.

There are five orbitals in a 6d subshell

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