Kaplan Exam FL 7 C/P Q17

[Pre-Med Oso]

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Hello there, I am trying to see how they converted to Angstrom. I know we have to use the following formula E = -Rh ( 1/n(initial)^2 - 1/n(final)^2 ( Rh = 2.18x10^-18). Does someone know the energy of a photon to Angstrom conversion it would help a lot thank you.


When an electron falls from n=3 to n=2 in a hydrogen atom, what is the value of the energy released, given that A is the energy needed to move an electron from the ground state of hydrogen atom to an infinite distance from the atom?

A: 0 A
B: 0.14A
C: 1.66A
D: 5.75A

B= Answer

 

[AptarePrep]

Hello there, I am trying to see how they converted to Angstrom. I know we have to use the following formula E = -Rh ( 1/n(initial)^2 - 1/n(final)^2 ( Rh = 2.18x10^-18). Does someone know the energy of a photon to Angstrom conversion it would help a lot thank you.


When an electron falls from n=3 to n=2 in a hydrogen atom, what is the value of the energy released, given that A is the energy needed to move an electron from the ground state of hydrogen atom to an infinite distance from the atom?

A: 0 A
B: 0.14A
C: 1.66A
D: 5.75A

B= Answer

Hi. I don't think A is Angstrom here. They are just using A as a variable for the value of the energy needed to move an electron from the ground state of hydrogen atom to an infinite distance from the atom. The value of A is the Rh value of 2.18x10^-18.
Therefore, when you solve for E for going from level 3 to level 2, you should get 3.0x10e-19.

Now, they just want a ratio of E to A:
3.0x10e-19 / 2.18x10e-18 = 0.14

Hence, the ratio of E to A is
0.14A