Question

Answers for questions 1-12 please

The energy level diagram for a hydrogen atom is shown. The following 000 ev questions are about the energy levels of the hydrogen atom. An0544 ev -1.51 ev 0.850 ev electron jumps from the n 5 level to the n 1 level. 1. Will this result in an emission line, or an absorption line in then-3 n 4 spectrum of this atom? 2. Which excited state did the electron start at? n-2 3.40 ev 2. What is the principal quantum number of the electron when it is at the level n = 1? 3. 3. n 1 -136 ev 4. Which series does this jump belong to? 5. What is the energy, in Joules, of the energy packet or particle associated with this jump? 5. 6. What is the frequency of this energy packet? 6. 7. What is the wavelength, in nanometers (nm) of the photon associated with this jump? A photon having an energy of 10.2 ev now strikes this hydrogen atom. Can the photon be absorbed by the atom? If so, describe the electron jump that will occur. If not, explain why not. 8. 8. 9. 9. What is the frequency of this photon? 10 10. What is the wavelength in nanometers (nm) of this photon? The following questions are about nuclear isotopes. 11. Explain what isotopes are. How are the isotopes of a particular element similar? How are they different? 12. An element has a mass number of 38 and is the 15th element on the periodic table. How many 12. neutrons does it have

Answer 1

1) Emission spectrum

When an electron jumps from higher energy level to lower, it emits energy.

2) The electron started at the fourth excited state

3) 1

'n' is the principle quantum number. Here, n=1, therefore, the principle quantum number of the electron is one.

4) Lyman series

5) Here, the energy levels are n=5 and n=1. Si the Rydberg's equation will allow you to calculate the wavelength of the photon emitted by the electron during the transition.

  $\frac{1}{\lambda}=R(\frac{1}{n_2^2}-\frac{1}{n_1^2})\\\\Here,\,\, n_1 = 5\,\, n_2=1$

where $\lambda$ is the wavelength and R is the Rydberg's constant which is equal to $1.0974\times10^7 \,\,m^-^1$

So we have to solve for wavelength.

  $\frac{1}{\lambda}=1.0974\times10^7(\frac{1}{1^2}-\frac{1}{5^2})$

$=1.05\times10^7 \,\,m^-^1$

$\Rightarrow\lambda=\frac{1}{1.05\times10^7 \,\,m^-^1}=9.49\times 10^-^8\,\,m=94.9\times 10^-^9 m=95 \,nm$

Since, energy $E=\frac{hc}{\lambda}$

where h is the Planks constant and c is the velocity of light.

$h=6.626\times10^-^3^4 J/s\\c=3.8\times10^8 m/s$

Energy $E=\frac{6.626\times10^-^34\times 3\times10^8}{9.49\times 10^-^8}=2.09\times 10^-^34 \,\, J$

6) Frequency, $f=\frac{c}{\lambda}=\frac{3\times10^8}{9.49\times 10^-^8}=3.16\times 10^1^5\,\,Hz$

7) Wavelength,$\lambda= 95\,\,nm$

9) Energy ,$E=\frac{hc}{\lambda}=10.2 eV$$=10.2\times1.602\times 10^-^1^9 \,\,J = 1.634\times10^-^1^8 \,\, J$

Wavelength, $\lambda=\frac{hc}{E}=\frac{6.626\times10^-^34\times3\times10^8}{1.634\times10^-18}=1.2\times 10^-^7 m$ $frequency, f=\frac{c}{\lambda}=\frac{3\times10^8}{1.2\times10^-^7}=2.47\times10^15 \,\, Hz$

10) Wavelength,$\lambda=1.22\times 10^-^7 m=122\times 10^-^9m=122 \,\,nm$

​​​​11) Isotopes are variants of a particular chemical element, which differ in neutron number and consequently in nucleon number.

All isotopes of a given element have the same number of protons but different numbers of neutrons in each atom.

12) Mass number = number of protons + number of neutrons= 38

Atomic number =number of protons/electrons=15

Number of neutrons = mass number - atomic number = 38 - 15 = 23

So the element has 23 neutrons.

Answer 2