Calculate the wovelength of light that is emitted or absorbed in the following e

Question

Calculate the wovelength of light that is emitted or absorbed in the following electron transitions in the hydrogen atom. Clearly state whether energy h being emitted or absorbed. What type of electromagnetic radiation (UV, visible, etc) h being emitted/absorbed for each ? a) From n=5 to n=2 b) From n=2 to n=5 c) From n=l to n=7 d) From n= 1 fo n= infinity 6. Which transition would represent the ionization of hydrogen (complete loss of its electron) Use the bohr equation to calculate the amount of energy (in joules) required to ionize a hydrogen atom,

Explanation / Answer

5.

a)

from, n=5 to n=2

Using the Rydberg equation,

1 / lambda = R (1/nf^2 - 1/ni^2)

1 / lambda = (1.097 x 10^7 m^-1)(1/(5^2) - 1/(2^2))

1 / lambda = (1.097 x 10^7 m^-1)(0.04-0.25)

1 / lambda = - 2.30 x 10^6 m^-1
lambda =- 1 / (2.30 x 10^6 m)

1lambda = - 4.35 x 10^-7 m = 435 nm .

Here – sign means energy is emitted.

The range of visible is 400 -700 nm thus it comes in visible.

b)                                

from, n=2 to n=5

Using the Rydberg equation,

1 / lambda = R (1/nf^2 - 1/ni^2)

1 / lambda = (1.097 x 10^7 m^-1)(1/(2^2) - 1/(5^2))

1 / lambda = (1.097 x 10^7 m^-1)(0.25-0.04)

1 / lambda = 2.30 x 10^6 m^-1
lambda = 1 / (2.30 x 10^6 m)

1lambda = 4.35 x 10^-7 m = 435 nm .

Here + sign means energy is absorbed.

The range of visible is 400 -700 nm thus it comes in visible.

c)

from, n=1 to n=7

Using the Rydberg equation,

1 / lambda = R (1/nf^2 - 1/ni^2)

1 / lambda = (1.097 x 10^7 m^-1)(1/(7^2) - 1/(1^2))

1 / lambda = (1.097 x 10^7 m^-1)(0.02 -1.0)

1 / lambda = - 1.07x 10^7 m^-1
lambda =- 1 / (107 x 10^7 m)

1lambda = -9.35 x 10^-8 m = 93.45 nm .

Here - sign means energy is emitted.

it comes in UV .

d)

from, n=1 to n=

Using the Rydberg equation,

1 / lambda = R (1/nf^2 - 1/ni^2)

1 / lambda = (1.097 x 10^7 m^-1)(1/() - 1/(1^2))

1 / lambda = (1.097 x 10^7 m^-1)(0.0 -1.0)

1 / lambda = - 1.097x 10^7 m^-1
lambda =- 1 / (1097 x 10^7 m)

1lambda = -9.11 x 10^-8 m = -93.11 nm .

Here - sign means energy is emitted

The above wavelength comes in UV.

6.

Complete loss of H electron means the transition from n=1 to n=, the wave length is calculate as follows:

from, n=1 to n=

Using the Rydberg equation,

1 / lambda = R (1/nf^2 - 1/ni^2)

1 / lambda = (1.097 x 10^7 m^-1)(1/() - 1/(1^2))

1 / lambda = (1.097 x 10^7 m^-1)(0.0 -1.0)

1 / lambda = - 1.097x 10^7 m^-1
lambda =- 1 / (1097 x 10^7 m)

1lambda = -9.11 x 10^-8 m = -93.11 nm .

Here - sign means energy is emitted

The above wavelength comes in UV.

To calculate the energy use following exprre4ssion:

E = c /

here
E = energy per photon
= plancks constant = 6.626x10^-34 J s
c = speed of light = 3.00x10^8 m/s
= wavelength ; 9.11 x 10^-8 m

E = c /
E= 6.626x10^-34 J s *3.00x10^8 m/s /9.11 x 10^-8 m

E= 2.18x 10^-18 J

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