What is the wavelength of the line corresponding to n=4 in the balmer series?

R= 1.097*10^7 m^-1

A) What is the wavelength of the line corresponding to n=4 in the
Balmer series?

B) What is the wavelength of the line corresponding to n=5 in the
Balmer series?

C) What is the smallest wavelength in the Balmer's series?

D) What is the largest wavelength in the Balmer series?

E) What is the smallest value of n for which the wavelength of a
Balmer series line is smaller than 400 nm, which is the lower limit
for wavelengths in the visible spectrum?

F) If m=1, in what range are the wavelengths calculated from the
generalized formula shown above?

a) microwave (1 to 10^-4 m)
b) infrared (10^-3 to 7*10^-7 m)
c) visible (7*10^-7 to 4*10^-7 m)
d) ultraviolet (4*10^-7 to 10^-8 m)
e) X rays (10^-8 to 10^-13 m)

G) If m=3, in what range are wavelengths calculated from the
generalized formula shown above?

a) microwave (1 to 10^-4 m)
b) infrared (10^-3 to 7*10^-7 m)
c) visible (7*10^-7 to 4*10^-7 m)
d) ultraviolet (4*10^-7 to 10^-8m)
e) X rays (10^-8 to 10^-13 m)

Answer

General guidance

Concepts and reason

The concept required to solve this problem is Balmer series.

First, calculate the wavelength using the formula for the Balmer series of the spectral line for different values of n. After that, calculate the smallest wavelength in the Balmer’s series by using the formula for Balmer series. Later, calculate the largest wavelength in the Balmer’s series by using the formula for Balmer series. Later, calculate the smallest value of n, which is the lower limit for wavelength in the visible spectrum by using the formula for Balmer series. Finally, calculate the range for the wavelengths for the corresponding values of m by using the generalized formula for Balmer series.

Fundamentals

Expression for the Balmer series to find the wavelength of the spectral line is as follows:

Ι
Ι)I

Here, is the wavelength, R is the Rydberg constant, and n is an integral value.

Step-by-step

Step 1 of 7

(A)

Expression for the Balmer series to find the wavelength of the spectral line is as follows:

Ι
Ι)I

Substitute 1.097x10m for R and 4 for n in the above equation.

=(1.097x10 mਹਿ ਤੇ
= 0.20568x10 me!
A =4.86x10 m

Part A

The wavelength of the spectral line is 4.86x10-m.

Balmer series is the series of lines of the atomic hydrogen in the visible and the ultraviolet spectrum. The wavelength of the photon emitted in this series corresponds to the electronic transition at the energy level 4 is in visible range of wavelengths.

[Common mistake]

Do not us the relation to calculate the wavelength. It is a mistake because the correct formula is Ι
Ι)I.

Step 2 of 7

(B)

Expression for the Balmer series to find the wavelength of the spectral line is as follows:

Ι
Ι)I

Substitute 1.097x10m for R and 5 for n in the above equation.

=(1.097x10 m)
ਤੇ
= 0.22304x10 ml
A =4.34x10 m

Part B

The wavelength of the spectral line is 4.34x10-m.

Balmer series is the series of lines of the atomic hydrogen in the visible and the ultraviolet spectrum. The wavelength of the photon emitted in this series corresponds to the electronic transition at the energy level 5 is in visible range of wavelengths.

[Common mistake]

Do not us the relation to calculate the wavelength. It is a mistake because the correct formula is Ι
Ι)I.

Step 3 of 7

(C)

Expression for the shortest wavelength of the Balmer series is,

(주)

Substitute 1.097x10m for R and for n in the above expression.

L = (1.097x10 m)( 7+
=0.274x10 m
Ramin = 3.647x10-m

Part C

The smallest wavelength of the spectral line is 3.647x10-7 m.

The shortest wavelength in the Balmer series for the resultant photon is due to the electronic transition from infinite to the second line of Balmer series. If the electron jumps between the greatest number of levels, then the largest amount of energy is released. Thus, the levels can be taken from infinity to 2. Energy is inversely proportional to the wavelength. Therefore, wavelength is smallest.

[Common mistake]

Do not us the relation to calculate the wavelength. It is a mistake because the correct formula is .

Step 4 of 7

(D)

Expression for the largest wavelength of the Balmer series is,

(주)

Substitute 1.097x10m for R and 3 for n in the above expression.

<= (1.97x10 m) ( 11
= 0.15236x10m
= 6.563 x 10-7 m
2

Part D

The largest wavelength of the spectral line is 6.536x10-7 m.

The longest wavelength in the Balmer series for the resultant photon is due to the electronic transition from third to second line of Balmer series. If the electron jumps between the lowest levels, then less amount of energy is released. Thus, the levels can be taken from 3 to 2. Energy is inversely proportional to the wavelength. Therefore, wavelength is largest.

Step 5 of 7

(E)

Expression for the smallest value of n for the wavelength of Balmer series, which is smaller than 400 nm is,

Ι
Ι)I

Substitute 1.097x10m for R and 400 nm for in the above equation.

LU
DU
8770-
szxo=(67
1w op 20)/(wat ook) + (6)
( :9). w.0x201) – u op

Part E

The smallest value of n is 7.

In Balmer series, a photon is emitted due to the transition between one state to another state. The smallest value of n is 7 and the resultant wavelength is smaller than 400 nm. If the value of n is taken as 6 then the resultant wavelength is greater than 400 nm and it is in between the visible spectrum.

Step 6 of 7

(F)

Expression for the Balmer series to find the wavelength of the spectral line is as follows:

(주)

Substitute 1.097x10m for R and 1 for m in the above equation.

1 = (1.097x10? m)( 7
)
2=1.215x10 m

Part F

The required range for the wavelength is 7x10-7m to 1x10 m which is in ultraviolet region.

If m=1, then the energy gap between the Balmer series of lines will be increase. Hence, the resultant wavelength is in the range of ultraviolet region.

Step 7 of 7

(G)

Expression for the Balmer series to find the wavelength of the spectral line is as follows:

(주)

Substitute 1.097x10m for R and 3 for m in the above equation.

<= (1.97x10 m) ( 11
= 0.15236x10m
= 6.563 x 10-7 m
2

Part G

The required range for the wavelength is 7x10-7m to 4x10-m which is in visible region.

If m=3, then the resultant range of the wavelength lies beyond the ultraviolet region. So, the resultant wavelength is in the range of visible region.

Answer

Part A

The wavelength of the spectral line is 4.86x10-m.

Part B

The wavelength of the spectral line is 4.34x10-m.

Part C

The smallest wavelength of the spectral line is 3.647x10-7 m.

Part D

The largest wavelength of the spectral line is 6.536x10-7 m.

Part E

The smallest value of n is 7.

Part F

The required range for the wavelength is 7x10-7m to 1x10 m which is in ultraviolet region.

Part G

The required range for the wavelength is 7x10-7m to 4x10-m which is in visible region.

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