Calculate the wavelength of light emitted when each of the following transitions occur in the hydrogen atom. what type of electromagnetic radiation is emitted in each transition? a. n = 4 → n = 3b. n = 5 → n = 4c. n = 5 → n = 3

a. 1875 nm

b. 4051 nm

c. 1282 nm

These all are infrared electromagnetic radiation.

Explanation:

Our strategy here is to utilize the Rydberg equation for hydrogen atom electronic transition.

1/λ = Rh x (1/n₁² - 1/n₂²)   where  λ is the wavelength

                                                   Rh is Rydberg constant

                                                   n₁ and n ₂ are the energy levels ( n₁ < n₂ )

Now lets star the calculations.

a.  n₁  = 3, n₂ = 4

1/λ = 1.097 x 10⁷ /m x (1/3² - 1/4²) = 5.333 x 10⁵/m

λ  = 1/(5.333 x 10⁵ /m) = 1.875 x 10⁻⁶ m

Converting λ to nanometers:

1.875 x 10⁻⁶ m x (1 x10⁹ nm/m) = 1875 nm

b.  n₁  = 4, n₂ = 5

1/λ = 1.097 x 10⁷ /m x  (1/4² - 1/5²) = 2.468 x 10⁵/m

λ  = 1/(2.468 x 10⁵/m) = 4.051 x 10⁻⁶ m

4.051 x 10⁻⁶ m x  (1 x10⁹ nm/m)  = 4051 nm

c.  n₁  = 3, n₂ = 5

1/λ = 1.097 x 10⁷ /m x  (1/3² - 1/5²) = 7801 x 10⁵/m

λ  = 1/(7801 x 10⁵/m) = 1282 x 10⁻⁶ m

1282 x 10⁻⁶ m x  (1 x10⁹ nm/m)  = 1282 nm

All of these transitions fall in the infrared region of the spectrum.

question: ray weighs 100 kilograms, his wrestling partner, bobby, weighs 50 kilograms. if ray moves at 4 m s , how fast will bobby have to run at him to apply an equal amount of kinetic energy?

answer: 8 m/s he needs to run twice as fast to keep up so he has some work to do

explanation:   if the mass has units of kilograms and the velocity of meters per second, the kinetic energy has units of kilograms-meters squared per second squared.

question answered by

(jacemorris04)

1.43 x 10^-11 j the guy commented the awnser just making sure u see it .