1. the emission spectrum of mercury atoms has a bright green line with wavelength 486.1 nm. 1nm = 1×10⁻⁹m a…

Explanation:

Step1: Recall the speed - wavelength - frequency relation

The speed of light $c = 3\times10^{8}\ m/s$, and the relation between speed of light $c$, wavelength $\lambda$ and frequency $\nu$ is $c=\lambda\nu$, so $\nu=\frac{c}{\lambda}$. Given $\lambda = 486.1\ nm=486.1\times 10^{-9}\ m$. $\nu=\frac{3\times 10^{8}}{486.1\times 10^{-9}}=\frac{3\times 10^{8}}{4.861\times 10^{-7}}\approx6.17\times 10^{14}\ Hz$

Step2: Calculate the energy of one photon

The energy of a photon is given by $E = h\nu$, and since $\nu=\frac{c}{\lambda}$, we can also write $E=\frac{hc}{\lambda}$, where $h = 6.63\times 10^{-34}\ J\cdot s$. $E=\frac{(6.63\times 10^{-34})\times(3\times 10^{8})}{486.1\times 10^{-9}}=\frac{19.89\times 10^{-26}}{4.861\times 10^{-7}}\approx4.09\times 10^{-19}\ J$

Step3: Calculate the energy of one - mole of photons

One mole of photons contains $N_A = 6.022\times 10^{23}$ photons. The energy of one - mole of photons $E_{mol}$ is $E_{mol}=N_A\times E$. $E_{mol}=(6.022\times 10^{23})\times(4.09\times 10^{-19})=6.022\times4.09\times 10^{4}\approx2.46\times 10^{5}\ J$

Answer:

a. $6.17\times 10^{14}\ Hz$ b. $4.09\times 10^{-19}\ J$ c. $2.46\times 10^{5}\ J$