Question

QUESTIONS Determine the end (final) value of n in the hydrogen atom transition, if electron starts in n-4 and the atom emits a photon of light with a wavelength of 486.

Answer 1

We must find the energy of photon emitted using its wavelength, and then we can find the final energy level, n_f.

Given,

$\lambda$=486 nm = 486 * 10^-9 m

We know,

Energy of photon emitted, E = hc / $\lambda$

where,

h =Plank's constant = 6.626*10^-34 J s

c= speed of light(photon) = 3*10^-8 m s^-1

Therefore,

E = hc / $\lambda$

=(6.626*10^-34 J s) (3*10⁸ m s^-1) / (486 * 10^-9 m)

= 4.09 *10 ^-19 J

Now, we will use the equation

E = R_H [(1/ni²)- (1/nf)²]

where,

R_H = Rydberg's constant =2.18 *10^-18 J

ni = initial energy state

nf = final energy state

Given,

ni =4

Plugging all the values in the above equation to determine nf, we get

E = R_H [(1/ni²)- (1/nf)²]

4.09 *10 ^-19 J=(2.18 *10^-18 J) [(1/4²)- (1/nf)²]

On solving for nf, we get

nf= 2

Therefore, the final value of n =2