Question

An electron in the n1 = 6 energy level of an H atom drops to a lower energy level; the atom emits a photon of wavelength 410. nm. To what energy level did the electron move?

Answer 1

Answer:

n = 2

Explanation:

As we know that when energy level of hydrogen atom goes down then the difference amount of energy is lost in term of photons

Now the wavelength of photons released is given as

$\lambda = 410 nm$

energy released is given by

$E = \frac{hc}{\lambda}$

now we have

$E = \frac{(6.6 \times 10^{-34})(3 \times 10^8)}{410 \times 10^{-9}}$

here we have

$E = 3.02 eV$

now we know that energy of n = 6 state

$E = -13.6\frac{1}{6^2} = -0.38 eV$

so we will have

$\Delta E = E_2 - E_1$

$3.02 eV = (-0.38) - E$

$E = -3.4 eV$

now we know that

$-3.4 = -13.6\frac{1}{n^2}$

$n = 2$