A hydrogen atom in an initial state having an ionization energy (the energy required to remove the electron from the hydrogen atom) of 0.85 eV makes a direct transition to a final state with an excitation energy (the difference in energy between the state and the ground state) of 10.2 eV. Required:
a. Calculate the energy (in eV) of the emitted photon during the transition described above.
b. Consider all possible transitions from the initial state and final state of the hydrogen atom described above.

Explanation:

a )

Energy of electron in fourth orbit of hydrogen atom = - 13.6 eV/ 4²

= -  .85 eV .

Energy of ground state  ( n = 1 ) = - 13.6 eV

energy of orbit n = 2

= 13.4 eV / 2² = -3.4 eV

difference = - 13.6 - ( - 3.4 eV)

= 10.2 eV

So transition is from 4 th orbit to 2 nd orbit

Energy difference = - 0.85 eV - ( - 3.4 eV )

= 2.55 eV

Energy of emitted photon = 2.55 eV .

b )

Possible transitions

1 ) from n = 4 to n = 3

2 ) from n = 3 to n = 2

3 ) from n = 4 to n = 2 .

found in the nucleus

has mass of one amu

answer: at 24 february, 2014 the population gets twice of its initial population.

explanation:

since we have given that

the population for alpha city, t years after january 1, 2004 is given as

first we find out the initial population, i.e. t=0,

so, our quadratic equation becomes,

according to question, we have said that if the population has twice its initial population, then it becomes

so, time taken to reach this above mentioned population is given by

now, we know that time can't be negative so, we take t=9.149 years.

hence, at 24 february, 2014 the population gets twice of its initial population.