Question

A hydrogen atom transitions from the n = 6 excited state to the n = 3 excited state, emitting a photon. a) What is the energy, in electron volts, of the electron in the n = 6 state? How far from the nucleus is the electron? b) What is the energy, in electron volts, of the photon emitted by the hydrogen atom? What is the wavelength of this photon? c) How many different possible photons could the n = 6 electron generate by dropping to a lower level (not just the n = 3 state)? d) After reaching the n = 3 state, the electron absorbs an photon with an energy of 50 eV, which ionizes the hydrogen atom (i.e., the electron reaches n=infinity). What is the wavelength of the electron after it is ejected from the hydrogen ato

Answer 1

Step-by-step explanation:

a ) energy, in electron volts, of the electron in the n = 6 state

= -13.6 / 6² eV

= - .378 eV .

b )

energy of n=3

= - 13.6 / 3²

= - 1.5111 eV

Difference = - .378 + 1.511

= 1.133 eV

the energy, in electron volts, of the photon emitted by the hydrogen atom

= 1.133 eV

c ) No of possible photons

= 6C₂

= 15

d) energy at n = 3

= - 1.5111 eV

50 eV energy is added to it so its energy

= 50 - 1.5111 eV

= 48.4889 eV .

From De broglie wavelength formula

λ = h / √ mE , m is mass of electron , E is kinetic energy h is plank's constant.

λ = 6.6 x 10⁻³⁴ / √ ( 9.1 x 10⁻³¹ x 48.4889 x 1.6 x 10⁻¹⁹ )

= 6.6 x 10⁻³⁴ / 26.57 x 10⁻²⁵

= .248 x 10⁻⁹ m

= .248 nm .