Question

Question 1

The lines observed in H-lamp are parts of Balmer Series. All these lines are due to a transition from various higher n value to a common n. What is this common lower n value for all these lines?

0

1

2

3

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Question 2

The bright violet line of Hg lamp has a wavelength of 435.8 nm. What is the energy of the photon associated with this emission line?

4.56*10^-19 J

4.56*10^-28 J

6.88*10^14 J

2.89*10^-31 J

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Question 3

Given: In Atomic Spectra lab, a student obtained his best-fit line equation to be y = 0.29 x + 46.8 when he plotted his Vernier reading on the y-axis and wavelength in nm on the x-axis.

Question: If the Vernier reading for a line is 185.4, what is the predicted wavelength of this line in nm? (Keep 4 sig figs with one decimal place.)

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Question 4

The only electron in a hydrogen atom moved from n=2 to n=6. What is the wavelength of photon (in nm) being associated with this transition? Use whole numbers of nm without any decimal places, such as "415", "1125".

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Question 5

Which of the following transitions in a hydrogen atom will absorb the highest amount of energy?

n = 5 to n = 1

n = 1 to n = 2

n = 2 to n = 100

n = 10 to n = 10000

Answer 1

1).

In layman series electron goes upto principle number n = 1

In this case n = 1 is the final for layman series.

So the answer is 1

2).

Energy (E) = hc/l

h is planks constant = 6.626 E-34 Js ,c is speed of light = 3.0 E8 m/s

lets convert l into m

l= 435.8 nm x 1 E-9 m / 1 nm

= 4.36 E-7 m

Lets put these value to get E

E = 6.626E-34 Js x 3.0 E8 m/s / 4.36 E-7 m

= 4.56E-19 J

Energy per photon = 4.56 E-19 J

Question 3

Equation

y = 0.29 x + 46.8

185.4 = 0.29 x + 46.8

Lets find x

0.29 x = 138.6

x = 477.9 nm

wavelength = 477.9 nm

Question 4

Lets use Rydberg equation to calculate lambda

1/l= R_H ( 1/ 2² – 1/6² )

R_H is Rydberg constant = 1.097 E7 per m

Lets plug given values

1/ l= 1.097 E7 per m x ( ¼ - 1/36 )

l= 4.10 E-7 m

l in nm

= 1.40 E-7 x 1E9 nm / 1m

= 410 nm

Question 5 )

The final value should be large in order to get higher number in the bracket.

In given n=10 and n = 10000 the bracket value will higher as compare to other so the pair n = 10 (initial) and n =10000 final will absorb higher energy.