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Show that of hydrogen’s spectral series—Lyman, Balmer, Paschen, and so on—only the four Balmer lines of Section 3 are in visible range (400-700nm)

Short Answer

Expert verified

It is shown that only the first four lines of the Balmer series are in the visible range, that is between 400-700nm.

Step by step solution

01

Given data

The frequency range of the visible spectrum is 400-700nm.

02

Wavelength of Hydrogen lines

The wavelength of emission when electron jumps from orbit n1 to n2orbit is

role="math" localid="1659966880476" λ=R-1(1n12-1n22)-1.....(I)

Here R is the Rydberg constant of value

R=1.097×107m-1

03

Determining the wavelength of Hydrogen lines in the visible range

For Lyman series n1=1. The maximum wavelength obtained from Lyman series is when n2=2the value of which from equation (I) is

λ=1.097×107m- 1-1112-122-1=121.5nm

This is less than the visible range. So Lyman series is not in the visible range.

For Paschen series n1=3. The minimum wavelength obtained from Paschen series is when n2=the value of which from equation (I) is

λ=1.097×107m- 1-1132-1-1=820.4nm

This is greater than the visible range. So Paschen series is not in the visible range. Thus only the Balmer series can be in the visible range.

For Balmer series n1=2. The wavelength of the first line obtained from the Balmer series is when n2=3the value of which from equation (I) is

λ=1.097×107m- 1-1122-132-1=656.3nm

The wavelength of the second line obtained from the Balmer series is when n2=4the value of which from equation (I) is

λ=1.097×107m- 1-1122-142-1=486.2nm

The wavelength of the third line obtained from the Balmer series is when n2=5 the value of which from equation (I) is

λ=1.097×107m- 1-1122-152-1=434.1nm

The wavelength of the fourth line obtained from the Balmer series is when n2=6the value of which from equation (I) is

λ=1.097×107m- 1-1122-162-1=410.2nm

The wavelength of the fifth line obtained from the Balmer series is when n2=7the value of which from equation (I) is

λ=1.097×107m- 1-1122-172-1=397nm

Thus only the first four lines are in the visible range

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Most popular questions from this chapter

Question: Consider a cubic 3D infinite well of side length of L. There are 15 identical particles of mass m in the well, but for whatever reason, no more than two particles can have the same wave function. (a) What is the lowest possible total energy? (b) In this minimum total energy state, at what point(s) would the highest energy particle most likely be found? (Knowing no more than its energy, the highest energy particle might be in any of multiple wave functions open to it and with equal probability.)

  1. What are the initial and final energy levels for the third (i.e., third-longest wavelength) line in the Paschen series? (See Figure 7.5)
  2. Determine the wavelength of this line.

An electron in a hydrogen atom is in the (n,l,ml) = (2,1,0) state.

(a) Calculate the probability that it would be found within 60 degrees of z-axis, irrespective of radius.

(b) Calculate the probability that it would be found between r = 2a0 and r = 6a0, irrespective of angle.

(c) What is the probability that it would be found within 60 degrees of the z-axis and between r = 2a0 and r = 6a0?

A spherical infinite well has potential energy

U(r)={0r<a+r>a

Since this is a central force, we may use the Schrodinger equation in the form (7-30)-that is, just before the specific hydrogen atom potential energy is inserted. Show that the following is a solution

R(r)=Asinbrr

Now apply the appropriate boundary conditions. and in so doing, find the allowed angular momenta and energies for solutions of this form.

How many different 3d states are there? What physical property (as opposed to quantum number) distinguishes them, and what different values may this property assume?

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