A photon with a wavelength of 2.29 × 10^–7 meter strikes a mercury atom in the ground state. Calculate the energy, in joules, of this photon. [Show all work, including the equat…

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asked Oct 28, 2017 142k views

2 Answers

Andreu Ramos · Answer 1 · 2017-11-04T07:31:36+0000

Answer:

Energy of photon,
$E=8.64* 10^(-19)\ J$

Step-by-step explanation:

It is given that,

Wavelength of photon,
$\lambda=2.29* 10^(-7)\ m$

It strikes a mercury atom in the ground state. We have to find the energy of this photon. It can be calculated using below relation as :

$E=(hc)/(\lambda)$

Where

h is the Planck's constant

c is the speed of light

So,
$E=(6.6* 10^(-34)\ J-s* 3* 10^8\ m/s)/(2.29* 10^(-7)\ m)$

$E=8.64* 10^(-19)\ J$

Hence, the above value is the energy of this photon.