Question

What wavelength (in nm) of photon absorbed when an electron transitions from then = 2 to the n = 21 state?Using-2.998 for light

Answer 1

The wavelength of the photon is 367.87 nm

Step-by-step explanation

Given information

`Speed\text{ of light = 2.998 }*\text{ 10}^8\text{ m/s}`

To find the wavelength of the photon absorbed, we will need to find the change in energy in transition using the below formula

`E_n\text{ = -2.18 }*\text{ 10}^(-18)\text{ }((1)/(n^2))`

From the question, the energy states given are from n = 21 to n =2

So, we can calculate the value of energy below

`E_(21)\text{ - E}_2\text{ = }\Delta E`
`\begin{gathered} \Delta E\text{ = -2.18 }*\text{ 10}^(-18)\text{ }((1)/(21^2)-(1)/(2^2)) \\ \Delta E\text{ = -2.18 }*10^(-18)((1)/(441)-(1)/(4)) \\ \Delta E\text{ = -2.18}*10^(-18)(0.0022675737\text{ - 0.25}) \\ \Delta E\text{ = -2.18}*10^(-18)\text{ }(-0.2477324263) \\ \Delta E\text{ = -2.18}*10^(-18)*(-2.477324263*10^(-1)) \\ \Delta E\text{ = 2.18}*2.477324263*10^(-18-1) \\ \Delta E\text{ = 5.400}*10^(-19)J \end{gathered}`

Since we have gotten the value of the energy, hence, we can now calculate the wavelength of the photon using the below formula

`\begin{gathered} \Delta E=(hc)/(\lambda) \\ Where \\ h\text{ = Planck's constant} \\ c\text{ = speed of light} \\ \lambda\text{ = wavelength} \end{gathered}`

Recall,

h = 6.626 x 10^-34 J.s

c = 2.998 x 10^8 m/s

`\begin{gathered} \lambda=(hc)/(\Delta E) \\ \frac{}{} \\ \lambda=(6.626*10^(-34)*2.998*10^8)/(5.40*10^(-19)) \\ \lambda=(6.626*2.998*10^(-34+8))/(5.40*10^(-19)) \\ \\ \lambda=(19.865*10^(-26+19))/(5.400) \\ \lambda=(19.865)/(5.40)*10^(-26+19) \\ \lambda=3.6787\text{ }*10^(-7) \\ \lambda=\text{ 367.87nm} \end{gathered}`

Hence, the wavelength of the photon is 367.87 nm