Question

. A bright violet line occurs at 435.8 nm in the emission spectrum of mercury vapor. What amount of energy, in joules, must be released by an electron in a mercury atom to produce a photon of this light?

Answer 1

Answer :  The energy released by an electron in a mercury atom to produce a photon of this light must be, $4.56\times 10^{-19}J$

Explanation : Given,

Wavelength = $435.8nm=435.8\times 10^{-9}m$

conversion used : $1nm=10^{-9}m$

Formula used :

$E=h\times \nu$

As, $\nu=\frac{c}{\lambda}$

So, $E=h\times \frac{c}{\lambda}$

where,

$\nu$ = frequency

h = Planck's constant = $6.626\times 10^{-34}Js$

$\lambda$ = wavelength = $435.8\times 10^{-9}m$

c = speed of light = $3\times 10^8m/s$

Now put all the given values in the above formula, we get:

$E=(6.626\times 10^{-34}Js)\times \frac{(3\times 10^{8}m/s)}{(435.8\times 10^{-9}m)}$

$E=4.56\times 10^{-19}J$

Therefore, the energy released by an electron in a mercury atom to produce a photon of this light must be, $4.56\times 10^{-19}J$