A system of 1610 particles, each of which is either an electron or a proton, has a net charge of −5.376×10^−17 C. How many electrons are in this system? What is the mass of this system?

Respuesta :

Answer:

Number of electrons in the system = 973.

Total mass of the system = [tex]\rm 1.064\times 10^{-24}\ kg.[/tex]

Explanation:

Assumptions:

  • [tex]\rm n_e[/tex] = number of electrons in the system.
  • [tex]\rm n_p[/tex] = number of protons in the system.
  • [tex]\rm q_e[/tex] = charge on an electron = [tex]\rm -1.6\times 10^{-19}\ C.[/tex]
  • [tex]\rm q_p[/tex] = charge on a proton = [tex]\rm +1.6\times 10^{-19}\ C.[/tex]
  • [tex]\rm m_e[/tex] = mass of an electron = [tex]\rm 9.11\times 10^{-31}\ kg.[/tex]
  • [tex]\rm m_p[/tex] = mass of a proton = [tex]\rm 1.67\times 10^{-27}\ kg.[/tex]

Given:

  • Total number of particles in the system, N = 1610.
  • Net charge on the system, q = [tex]\rm -5.376\times 10^{-17}\ C.[/tex]

Since, the system is comprised of electrons and protons only, therefore,

[tex]\rm N = n_e+n_p\\n_p=N-n_e\ \ \ \ \ \ ................\ (1).[/tex]

The net charge on the system can be written in terms of charges on electrons and protons as

[tex]\rm q=n_eq_e+n_pq_p\ \ \ \ \ ...................\ (2).[/tex]

Putting the value of (2) in (1), we get,

[tex]\rm q=n_eq_e+(N-n_e)q_p\\q=n_eq_e+Nq_p-n_eq_p\\q=n_e(q_e-q_p)+Nq_p\\n_e(q_e-q_p)=q-Nq_p\\n_e=\dfrac{q-Nq_p}{q_e-q_p}\\=\dfrac{-5.3756\times 10^{-17}-1610\times 1.6\times 10^{-19}}{-1.6\times 10^{-19}-1.6\times 10^{-19}}=972.98\\\Rightarrow n_e\approx 973\ electrons.[/tex]

It is the number of electrons in the system.

Therefore, the number of protons is given by

[tex]\rm n_p = N-n_e=1610-973=637.[/tex]

The total mass of the system is given by

[tex]\rm M=n_em_e+n_pm_p\\=(973\times 9.11\times 10^{-31})+(637\times 1.67\times 10^{-27})\\=1.064\times 10^{-24}\ kg.[/tex]