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Bohr Model of Hydrogen Atom

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Bohr Model of Hydrogen Atom Class 12th

Figure shows the energy levels P,Q,R,S and G of an atom where G is the ground state. A red line in the emission spectrum of the atom can be obtained by an energy level change from Q to S. A blue line can be obtained by following energy level change

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a month ago

Bohr Model of Hydrogen Atom Class 12th

The radius of inner most orbit of hydrogen atom is $5.3 * 10^{- 11} m$ . What is the radius of third allowed orbit of hydrogen atom?

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Raj Pandey

Contributor-Level 9

$r_{n} = a_{0} n_{2}$

$\begin{array}{r} r_{1} = a_{0} = 5.3 * 10^{- 11} m \\ r_{3} = a_{0} (3) 2 \\ = 5.3 * 10^{- 11} * 9 \\ = 4.77 ? \end{array}$

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2 months ago

Bohr Model of Hydrogen Atom Class 12th

Consider the following statements:
(A) Atoms of each element emit characteristics spectrum.
(B) According to Bohr's Postulate, an electron in a hydrogen atom, revolves in a certain stationary orbit.
(C) The density of nuclear matter depends on the size of the nucleus.
(D) A free neutron is stable but a free proton decay is possible.
(E) Radioactivity is an indication of the instability of nuclei.
Choose the correct answer from the options given below:

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Vishal Baghel

Contributor-Level 10

Density of nucleus is constant.

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2 months ago

Bohr Model of Hydrogen Atom Class 12th

In Bohr's atomic model, the electron is assumed to revolve in a circular orbit of radius 0.5 Å. If the speed of electron is 2.2 * 10? m/s, then the current associated with the electron will be _______ * 10?² mA. [Take π as 22/7]

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Vishal Baghel

Contributor-Level 10

r = 0.5Å = 0.5*10? ¹? m
v = 2.2*10? m/s
I =?
t = 2πr/v
I = e/t = ev/2πr = (1.6*10? ¹? * 2.2*10? )/ (2 * 22/7 * 0.5*10? ¹? )
I ≈ 1.12*10? ³ A = 1.12 mA = 112 * 10? ² mA

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2 months ago

Bohr Model of Hydrogen Atom Class 12th

According to Bohr atom model, in which of the following transitions will the frequency be maximum?

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Vishal Baghel

Contributor-Level 10

$Δ E = h v \to (\frac{1}{n_{2}^{2}} - \frac{1}{n_{1}^{2}})$

(1) n₁ = 3, n₂ = 2

$(\frac{1}{2^{2}} - \frac{1}{3^{2}}) = \frac{9 - 4}{36} = \frac{5}{36}$

(2) n₁ = 4, n₂ = 3

$(\frac{1}{3^{2}} - \frac{1}{4^{2}}) = \frac{16 - 9}{144} = \frac{7}{144}$

(3) n₁ = 2, n₂ = 1

$(\frac{1}{1^{2}} - \frac{1}{2^{2}}) = \frac{4 - 1}{4} = \frac{3}{4}$

(4) n₁ = 5, n₂ = 4

$(\frac{1}{4^{2}} - \frac{1}{5^{2}}) = \frac{25 - 16}{400} = \frac{9}{400}$

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