Answer:
(a) The energy of the photon is 1.632 x  J.
 J.
(b) The wavelength of the photon is 1.2 x  m.
 m.
(c) The frequency of the photon is 2.47 x  Hz.
 Hz.
Explanation:
Let;
 = -13.60 ev
 = -13.60 ev
 = -3.40 ev
 = -3.40 ev
(a) Energy of the emitted photon can be determined as;
 -
 -  = -3.40 - (-13.60)
 = -3.40 - (-13.60)
            = -3.40 + 13.60
            = 10.20 eV
            = 10.20(1.6 x  )
)
 -
 -  = 1.632 x
 = 1.632 x  Joules
 Joules
The energy of the emitted photon is 10.20 eV (or 1.632 x  Joules).
 Joules).
(b) The wavelength, λ, can be determined as;
E = (hc)/ λ
where: E is the energy of the photon, h is the Planck's constant (6.6 x  Js), c is the speed of light (3 x
 Js), c is the speed of light (3 x  m/s) and λ is the wavelength.
 m/s) and λ is the wavelength.
10.20(1.6 x  ) = (6.6 x
) = (6.6 x  * 3 x
 * 3 x  )/ λ
)/ λ
λ = 
   = 1.213 x 
Wavelength of the photon is 1.2 x  m.
 m.
(c) The frequency can be determined by;
E = hf
where f is the frequency of the photon.
1.632 x  = 6.6 x
  = 6.6 x  x f
 x f
f = 
  = 2.47 x  Hz
 Hz
Frequency of the emitted photon is 2.47 x  Hz.
 Hz.