Question

Find the frequency of light emitted in the transition from the 148th orbit to the 139th orbit. Find the frequency of light absorbed in the transition from the 165th orbit to the 174th orbit. Submit Answer

Answer 1

The frequency of light emitted in the transition from the 148th orbit to the 139th orbit is 2.00 × 10^15 Hz. For the absorption of light, use the same formula with the given wavelength.

Step-by-step explanation:

To calculate the frequency of light emitted in the transition from one orbit to another, we can use the formula:

frequency = (speed of light) / (wavelength)

First, we need to convert the given wavelength from nanometers to meters:

150 nm = 150 × 10-9 m

Then, we can plug this value into the formula:

frequency = (3.00 × 108 m/s) / (150 × 10-9 m) = 2.00 × 1015 Hz

Similarly, for the absorption of light, you can use the same formula with the given wavelength.

Answer 2

`f=3* 10^8* 64.90=19.4* 10^9 s^(-1)`

`f=3* 10^8*39.9537=119.8613* 10^8\ /sec`

Step-by-step explanation:

for the transition from 148th orbit to 138th orbit

we know the formula

`(1)/(\lambda )=R* \left ( (1)/(n^2_(final))-(1)/(n^2_(initial)) \right )`

where R is Rydberg constant whose value is
`1.0974* 10^7 /m`

putting this value in equation

`(1)/(\lambda )=1.0974* 10^7 \left ( (1)/(139^2)-(1)/(148^2) \right )`

`(1)/(\lambda )=64.9075`

`f=c\lambda`

`f=3* 10^8* 64.90=19.4* 10^9 s^(-1)`

for the transition from 165th to 174th orbit

we know the formula

`(1)/(\lambda )=R* \left ( (1)/(n^2_(final))-(1)/(n^2_(initial)) \right )`

where R is Rydberg constant whose value is
`1.0974* 10^7 /m`

putting this value in equation

`(1)/(\lambda )=1.0974* 10^7 \left ( (1)/(165^2)-(1)/(174^2) \right )`

`(1)/(\lambda )=39.95377`

`f=c\lambda`

`f=3* 10^8*39.9537=119.8613* 10^8\ /sec`