# What is the frequency and wavelength of a photon emitted during a transition from n = 6 to n = 1 state in the hydrogen atom ?

Mar 15, 2018

$v = - 3.2 \cdot {10}^{15} H z \mathmr{and} {s}^{-} 1$
λ=-9.4*10^(-8)m

#### Explanation:

Alright, given:
${n}_{i} = 6$
${n}_{f} = 1$
$z = 1$ z is the atomic number
${R}_{H} = 2.179 \times {10}^{- 18} J$ It is a constant

To find change in energy
∆E= -Z^2*R_H(1/n_f^2- 1/n_i^2)
∆E= -(1)^2*2.179xx10^(-18) J(1/(1)^2- 1/(6)^2)
∆E=-2.12*10^(-18)J

To Find frequency:
v= (∆E)/h
h is plank's constant: $6.626 \cdot {10}^{- 34} J s$
v= (∆E)/h
$v = \frac{- 2.12 \cdot {10}^{- 18} J}{6.626 \cdot {10}^{- 34} J s}$
$v = - 3.2 \cdot {10}^{15} H z \mathmr{and} {s}^{-} 1$

To find wavelength:
c/v= λ
Where c is the speed of light, $3 \cdot {10}^{8} \frac{m}{s}$
$\frac{3 \cdot {10}^{8} \frac{m}{s}}{- 3.2 \cdot {10}^{15} H z} = - 9.4 \cdot {10}^{- 8} m$