. 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 produ…

Question

asked Feb 18, 2021 121k views

1 Answer

Bishal Das · Answer 1 · 2021-02-21T19:48:51+0000

Answer : The energy released by an electron in a mercury atom to produce a photon of this light must be,
4.56* 10^(-19)J

Explanation : Given,

Wavelength =
435.8nm=435.8* 10^(-9)m

conversion used :
1nm=10^(-9)m

Formula used :

$E=h* \\u$

As,
$\\u=(c)/(\lambda)$

So,
$E=h* (c)/(\lambda)$

where,

$\\u$ = frequency

h = Planck's constant =
6.626* 10^(-34)Js

$\lambda$ = wavelength =
435.8* 10^(-9)m

c = speed of light =
3* 10^8m/s

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

E=(6.626* 10^(-34)Js)* ((3* 10^(8)m/s))/((435.8* 10^(-9)m))

E=4.56* 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* 10^(-19)J