Question

The ratio of speed of an electron in ground state in bohr's first orbit of hydrogen atom to velocity of light in air

Answer 1

orbital speed of electron is given by the following equation

$v = 2.2 * 10^6 \frac{z}{n}$

here z = atomic number

n = orbit of atom

now in the above case

if we need to find the speed of electron in first orbit of hydrogen atom

then z = 1 and n = 1

$v = 2.2 * 10^6 m/s$

speed of light in air is given by

$c = 3 * 10^8 m/s$

now to find the ratio we will have

$\frac{v}{c} = \frac{2.2* 10^6}{3*10^8}$

$\frac{v}{c} = 7.33 * 10^{-3}$

Answer 2

Answer:

Substituting the values, the values of m (mass of the electron), e (charge on the electron) and h (Planck’s constant), we get:

Substituting the values, the values of m (mass of the electron), e (charge on the electron) and h (Planck’s constant), we get: En=−21.8×10−19n2J/atom

Substituting the values, the values of m (mass of the electron), e (charge on the electron) and h (Planck’s constant), we get: En=−21.8×10−19n2J/atom⇒ −1312n2kJmole−1

Substituting the values, the values of m (mass of the electron), e (charge on the electron) and h (Planck’s constant), we get: En=−21.8×10−19n2J/atom⇒ −1312n2kJmole−1⇒ −13.6n2eV/atom (1eV=1.602 ×J)

Substituting the values, the values of m (mass of the electron), e (charge on the electron) and h (Planck’s constant), we get: En=−21.8×10−19n2J/atom⇒ −1312n2kJmole−1⇒ −13.6n2eV/atom (1eV=1.602 ×J)Where n=1,2,3, ----etc. stands for the 1st,2 Nd, 3rd ----etc. levels. The energy level which is closest to the nucleus has the lowest energy and the energy increases as the energy levels increase and so on.

Substituting the values, the values of m (mass of the electron), e (charge on the electron) and h (Planck’s constant), we get: En=−21.8×10−19n2J/atom⇒ −1312n2kJmole−1⇒ −13.6n2eV/atom (1eV=1.602 ×J)Where n=1,2,3, ----etc. stands for the 1st,2 Nd, 3rd ----etc. levels. The energy level which is closest to the nucleus has the lowest energy and the energy increases as the energy levels increase and so on.But for hydrogen like particles, the expression for energy is:

Substituting the values, the values of m (mass of the electron), e (charge on the electron) and h (Planck’s constant), we get: En=−21.8×10−19n2J/atom⇒ −1312n2kJmole−1⇒ −13.6n2eV/atom (1eV=1.602 ×J)Where n=1,2,3, ----etc. stands for the 1st,2 Nd, 3rd ----etc. levels. The energy level which is closest to the nucleus has the lowest energy and the energy increases as the energy levels increase and so on.But for hydrogen like particles, the expression for energy is: En=−2π2mZ2e4n2h2⇒ −1312Z2n2kJmole−1⇒ −2.18×10−11n2ergs