Question

imagine that germanium has an emission spectrum with a spectral line with a wavelength of 640 nm determine the frequency and energy of the spectral line.\\(\frac{3.00\times10^{8}}{6.40\times10^{-7}} = 0.46875 = 4.6875\times10^{14}\\ hz\\)\\(640\times10^{-9}=6.4\times1\\)

Explanation:

Step1: Convert wavelength to meters

Given wavelength $\lambda = 640\space nm$. Since $1\space nm = 10^{-9}\space m$, we have $\lambda = 640\times10^{-9}\space m = 6.4\times10^{-7}\space m$.

Step2: Calculate frequency ($

u$)
The relationship between speed of light ($c$), wavelength ($\lambda$) and frequency ($
u$) is $c = \lambda
u$. So, $
u=\frac{c}{\lambda}$.
We know $c = 3.00\times10^{8}\space m/s$ and $\lambda = 6.4\times10^{-7}\space m$.
Substituting the values, $
u=\frac{3.00\times10^{8}}{6.4\times10^{-7}}=\frac{3.00}{6.4}\times10^{8 + 7}\approx0.46875\times10^{15}=4.6875\times10^{14}\space Hz$.

Step3: Calculate energy ($E$)

The formula for energy of a photon is $E = h
u$, where $h = 6.626\times10^{-34}\space J\cdot s$ (Planck's constant) and $
u = 4.6875\times10^{14}\space Hz$.
Substituting the values, $E=(6.626\times10^{-34})\times(4.6875\times10^{14})$.
First, multiply the coefficients: $6.626\times4.6875\approx31.06$.
Then, multiply the powers of 10: $10^{-34}\times10^{14}=10^{-20}$.
So, $E\approx31.06\times10^{-20}\space J = 3.106\times10^{-19}\space J$. (We can also use $E=\frac{hc}{\lambda}$ directly. Substituting $h = 6.626\times10^{-34}\space J\cdot s$, $c = 3.00\times10^{8}\space m/s$ and $\lambda = 6.4\times10^{-7}\space m$: $E=\frac{6.626\times10^{-34}\times3.00\times10^{8}}{6.4\times10^{-7}}=\frac{19.878\times10^{-26}}{6.4\times10^{-7}}=\frac{19.878}{6.4}\times10^{-26 + 7}\approx3.106\times10^{-19}\space J$)

Answer:

Frequency: $\boldsymbol{4.69\times10^{14}\space Hz}$ (rounded to three significant figures)
Energy: $\boldsymbol{3.11\times10^{-19}\space J}$ (rounded to three significant figures)