The following four problems are about the following situation: A transition for

Question

The following four problems are about the following situation: A transition for an electron in the Bohr Hydrogen atom will be given. You determine what energy/frequency/wavelength photon is emitted.

1)  n=3 to n=1

a) 8.1 eV

b) 9.1 eV

c) 10.1 eV

d) 11.1 eV

e) 12.1 eV

2)  n=6 to n=1

a) 1.12 x 10-18 J

b) 2.12 x 10-18 J

c) 3.12 x 10-18 J

d) 4.12 x 10-18 J

e) 5.12 x 10-18 J

3)  n=4 to n=2

a) 186 nm

b) 286 nm

c) 386 nm

d) 486 nm

e) 586 nm

4)  n=10 to n=3

a) 1.11 x 1014 Hz

b) 2.22 x 1014 Hz

c) 3.33 x 1014 Hz

d) 4.44 x 1014 Hz

e) 5.55 x 1014 Hz

Explanation / Answer

1) energy in nth orbit is En = -13.6 /n^2

in n = 3 state

E3 = -13.6/3^2 = -1.51eV

in n= 1 state

E1 = -13.6/1 = -13.6 eV

energy emitted is E3-E1 = -1.51+13.6 = 12.1eV

So correct option is e) 12.1 eV

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2) n=6 to n=1

in n= 6

E6 = -13.6/6^2 =-0.377 eV

E1 = -13.6 eV

E6-E1 = -0.377+13.6 = 13.223 eV = 13.223*1.6*10^-19 = 2.12*10^-18 J

So here the corret option is b) 2.12*10^-18 J

======================================================

3) n= 4 to n=2

in n= 4

E4 = -13.6/4^2 = -0.85 eV

E2 = -13.6/2^2 = -3.4 eV

E4-E2 = -0.85+3.4 = 2.55 eV = 2.55*1.6*10^-9 = 4.08*10^-19 J

4.08*10^-19 = h*c/lamda

lamda = (h*c)/(4.08*10^-19)= (6.625*10^-34*3*10^8)/(4.08*10^-19) = 486 nm

So correct option is d) 486 nm

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4) n=10 to n=3

E10 = -13.6/10^2 = -0.136 eV

E3 = -13.6/3^3 = -13.6/9 = -1.51 eV

E10-E3 = -0.136+1.51 = 1.374 eV = 1.374*1.6*10^-19 = 2.2*10^-19 J

2.2*10^-19 = h*f

frequency f = (2.2*10^-19)/(6.625*10^-34) = 3.33*10^14 Hz

So the correct answer is c) 3.33*10^14 Hz

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