The Hydrogen Atom MCQ Quiz - Objective Question with Answer for The Hydrogen Atom - Download Free PDF

Calculation:
The wavelength is minimum for transitions having the maximum energy gap:

ΔEmax = Em+3 – Em = E15 – E12

ΔEmax = –0.544 – (–0.85) = 0.306 eV

Hence, the minimum wavelength is:

λmin = hc / ΔEmax = 1240 / 0.306 nm = 4052 nm

Smallest wavelength = 4052 nm

Calculation:

Six transitions are possible between four energy levels (k

The energy of the nth level is given by:

En = –13.6 Z² / n² eV

Thus, solving for the given energy values:

–13.6 Z² / m² = –0.85, (Equation 1)

–13.6 Z² / (m + 3)² = –0.544. (Equation 2)

By solving equations (1) and (2), we get Z = 3 and m = 12.

Correct option is: (2) h / 2πm

Magnetic moment

M = IA = I (πr²)

M = (ev / 2π) × (πr²)         ...(1)

Given B (πr²) = n (h / e)

⇒ r² = h / (Bπe)         ...(2)    (∵ n = 1)

And when charge is moving in external magnetic field

Then r = mv / qB

⇒ v / r = eB / m         ...(3)    (∵ q = e)

Put value from equation (2) and (3) in equation (1)

M = (ev / 2π) × (πr²)

M = (e / 2π) × (eB / m) × π × (h / Bπe)  = eh / 2πm

CONCEPT:

The velocity of the electron in the  n^th orbit of the hydrogen atom is written as;

\(v = \frac{2 \pi kZe^2}{nh}\)

Here we have Z as the atomic number, e is the electron, and h is Planck's constant.

CALCULATION:

As we know, the velocity of the electron in the  nth orbit of the hydrogen atom is written as;

\(v_n = \frac{2 \pi kZe^2}{nh}\)  ----(1)

Therefore, from equation (1),

 \(v_n \propto \frac{1}{n}\)

Hence option 4) is the correct answer.

\(\Rightarrow \lambda_{max}=\dfrac {4}{3R}\approx 1213\overset {o}{A}\)

and \(\dfrac {1}{\lambda_{min}}=R\left[\dfrac {1}{(1)^2}-\dfrac {1}{\infty}\right]\lambda_{min}=\dfrac {1}{R}\approx 910\overset{o}{A}\)

CONCEPT:

• When the electrons in the atoms are in the state other than the ground state then this is called the atom in an excited state.
• The lifetime of atoms in an excited state is the time duration in which the electrons remain in their excited state.
  • The lifetime of atoms in an excited state is an average lifetime derived from the decay probability.
• Excited-state lifetimes are typically in few nanoseconds, The closest answer is 10^-8 seconds. So option 1 is correct.

CONCEPT:

• Bohr’s Model: Niels Bohr by making 4 postulates solved the puzzle of hydrogen spectra.
  • The mass of the nucleus is very large compared to that of the electrons and almost the entire mass of the atom is concentrated in the nucleus and hence assumed to be infinite.
  • The electrons revolve around the nucleus in circular orbits.
  • The mass of the electron remains constant.
  • The radius around the orbit has some special values of the radius. In these stationary orbits, they do not radiate energy as expected from Maxwell’s laws.
  • The energy of each stationary orbits are fixed, electrons can jump from a higher orbit to a lower orbit by emitting a photon of radiation. Where Higher energy orbit – Lower energy orbit = (h c)/λ  
    • An electron can also jump from lower to higher energy by absorbing energy.
  • In stationary orbits, the angular momentum L of an electron is an integral multiple of (h/2π)
    • L = n × (h/2π)

EXPLANATION:

• From the above, it is clear that all the above conditions are true. Hence option 4 is correct.

CONCEPT:

• Magnetic dipole moment. It is a magnetic property of electric current loops or magnets.
• The quantity of magnetic dipole moment is equal to the amount of current flowing in the loop multiplied by the area that the loop encompasses.

Magnetic dipole moment (μ) = IA

where I is current and A is the area.

• Angular momentum: the quantity of a rotating body, which is the product of its moment of inertia and its angular velocity.

Angular momentum (L) = m v r

where m is the mass of the body, v is velocity and r is the distance from the rotating point.

From Bohr's Model of Atom:A

​angular momentum (L) = mvr = nh/2π

v = nh/2πmr

Time period T = 2πr/v

current I = q/T = e/T = e/(2πr/v) = \(enh \over 4\pi^2mr^2\)

where n is the orbit number, h is plank constant, m is the mass of electron, r is the radius of orbit, v is the velocity of the electron, e is the charge on one electron.

CALCULATION:

Magnetic dipole moment (μ) = IA

μ = \({enh \over 4\pi^2mr^2} \times A\) = \({enh \over 4\pi^2mr^2} \times \pi r^2\)

angular momentum (L) = nh/2π

\(\frac{μ}{L}=\frac{e}{2m}\)

So the correct answer is option 1.

CONCEPT:

• Hydrogen Spectrum and Spectral series: When a hydrogen atom is excited, it returns to its normal unexcited (or ground state) state by emitting the energy it had absorbed earlier.
• This energy is given out by the atom in the form of radiations of different wavelengths as the electron jumps down from a higher to a lower orbit.
• The transition from different orbits causes different wavelengths, these constitute spectral series which are characteristic of the atom emitting them.
• When observed through a spectroscope, these radiations are imaged as sharp and straight vertical lines of a single color.
  
• Mainly there are five series and each series is named after it's discovered as Lyman series (n1 = 1), Balmer series (n1​ = 2), Paschen series (n1​ = 3), Bracket series (n1​ = 4), and Pfund series (n1​ = 5).
• According to Bohr's theory, the wavelength of the radiations emitted from the hydrogen atom is given by
  \(\frac{1}{\lambda } = RZ^2\left[ {\frac{1}{{n_1^2}} - \frac{1}{{n_2^2}}} \right]\)

where

n2 = outer orbit (electron jumps from this orbit),

n1 = inner orbit (electron falls in this orbit),

Z = atomic number

R = Rydberg's constant.

CALCULATION:

Given-

For Lymen n1 = 2  and n2 = 3

And wavelength associated with electron jumping from any state to the first state according to Lyman series and it can be expressed as

 \(\frac{1}{\lambda }\; = \;R\left( {\frac{1}{{{2^2 }}} - \frac{1}{{{3^2}}}} \right)\)

​i.e., \(\lambda =\frac{1}{R\times (\frac{1}{{{2}^{2}}}-\frac{1}{{{3}^{2}}})}=\frac{1}{1.1\times {{10}^{7}}\times \frac{5}{4\times 9}}=6.545\times {{10}^{-7}}\approx 654nm\)

Concept:

From question, the atom jumps from one spectral line to another spectral line. The wavelength of the atom is given by the formula:

\(\frac{1}{{\rm{\lambda }}} = {{\rm{Z}}^2}{{\rm{R}}_\infty }\left( {\frac{1}{{{\rm{n}}_1^2}} - \frac{1}{{{\rm{n}}_2^2}}} \right)\)

Where,

‘Z’ is the atomic number

n₁ and n₂ is the principal quantum number

R_∞ is the Rydberg constant. (1.09737 × 10⁷ m^-1)

Calculation:

When the atom jumps from third excited state to the second excited state:

\( \Rightarrow \frac{1}{{{{\rm{\lambda }}_1}}} = {\left( 1 \right)^2}{{\rm{R}}_\infty }\left( {\frac{1}{{{3^2}}} - \frac{1}{{{4^2}}}} \right) \)

\( \Rightarrow \frac{1}{{{{\rm{\lambda }}_1}}} = {{\rm{R}}_\infty }\left( {\frac{1}{9} - \frac{1}{{16}}} \right)\)

\( \Rightarrow \frac{1}{{{{\rm{\lambda }}_1}}} = {{\rm{R}}_\infty }\left( {\frac{{16-9}}{{16 \times 9}}} \right)\)

\( \Rightarrow \frac{1}{{{{\rm{\lambda }}_1}}} = {{\rm{R}}_\infty }\left( {\frac{{7}}{{16 \times 9}}} \right)\)

When the atom jumps from second excited state to the first excited state:

\( \Rightarrow \frac{1}{{{{\rm{\lambda }}_2}}} = {\left( 1 \right)^2}{{\rm{R}}_\infty }\left( {\frac{1}{{{2^2}}} - \frac{1}{{{3^2}}}} \right)\)

\( \Rightarrow \frac{1}{{{{\rm{\lambda }}_2}}} = {{\rm{R}}_\infty }\left( {\frac{5}{36} } \right)\)

From question,
\( \Rightarrow \frac{{{{\rm{\lambda }}_1}}}{{{{\rm{\lambda }}_2}}} = \frac{{\frac{1}{{{{\rm{\lambda }}_2}}}}}{{\frac{1}{{{{\rm{\lambda }}_1}}}}}\)

\( \Rightarrow \frac{{{{\rm{\lambda }}_1}}}{{{{\rm{\lambda }}_2}}} = \frac{5/36}{(7/16\times9)}\)

Ans: Option: 1

CONCEPT: 

• Angular Momentum: The moment of momentum is called angular momentum.

  • It is the property of any rotating object given by moment of inertia times angular velocity.

• Bohr's postulate 2: In a hydrogen atom, the electron can revolve around the nucleus, without radiating energy, only in those orbits for which the angular momentum of the electron is equal to an integral multiple \(\frac{h}{2\pi }\) of where h is Planck's constant.

Angular momentum (J) = m v r = \(\frac{nh}{2\pi }\)

Where m is the mass of the electron, v is the velocity of the electron and r is the radius of the orbit

EXPLANATION:

Angular momentum (J) = m v r = \(\frac{nh}{2\pi }\)

by solving we get,  \(r= \frac{J}{mv}\)

CONCEPT:

• The radius of a hydrogen-like atom is given by:

r = a0n2/Z

where r is the radius of atom, Z is the atomic number of the atom, n is the orbit number, and a0 is the radius of the 1st orbit of hydrogen.

CALCULATION:

Given -  Z = 1, and n = 3

• The radius of the 1^st orbit of hydrogen atoms given by:

⇒ r₁ = a_o   

• The radius of the 3^rd orbits of hydrogen atoms given by:

⇒ r₃ = ao32/Z = a_o9       

⇒ r₃ = 9r               [∵ r = a₀] 

The correct answer is Nuclear Fusion.

Key Points

• Nuclear fusion
  • It is the nuclear reaction in which two or more atomic nuclei are combined to form one or more different atomic nuclei and subatomic particles.
  • Sun and other stars generate light and heat by nuclear fusion.
  • A hydrogen bomb is an immensely powerful bomb whose destructive power comes from the rapid release of energy during the nuclear fusion of isotopes of hydrogen (deuterium and tritium), using an atom bomb as a trigger.
  • The principle behind the hydrogen bomb is based on uncontrollable nuclear fusion. Hence, Option 3 is correct.
  • A nuclear bomb based on the fission of uranium is placed at the core of the hydrogen bomb.
  • So hydrogen bomb is based on the principle of Nuclear Fusion Reaction.
  • A hydrogen bomb is based on the principle of nuclear fusion.
• 

Important Points

• ​Nuclear fusion
  • It is the process where the nuclei of two light atoms combine to form a new nucleus.
  • A hydrogen bomb is considered to be 1,000 times more powerful than an atomic bomb.
  • Hydrogen bombs cause a bigger explosion.
  • As shock waves, blasts, heat, and radiation all have a larger reach than an atomic bomb.
• 

Additional Information

• A nuclear explosion
  • It is an explosion that occurs as a result of the rapid release of energy from a high-speed nuclear reaction. 
  • Atmospheric nuclear explosions are associated with mushroom clouds, although mushroom clouds can occur with large chemical explosions.
• A chain reaction
  • It refers to a process in which neutrons released in fission produce additional fission in at least one further nucleus.
  • If each neutron releases two more neutrons, then the number of fissions doubles each generation.

Concept:

Bohr's Atomic Model –

• Bohr proposed a model for hydrogen atom which is also applicable for some lighter atoms in which a single electron revolves around a stationary nucleus of positive charge Ze (called hydrogen-like atom).

Bohr's model is based on the following postulates –

• He postulated that an electron in an atom can move around the nucleus in certain circular stable orbits without emitting radiations.
• Bohr found that the magnitude of the electron's angular momentum is quantized
• i.e. \(L = mv\_n\;{r_n} = n\left( {\frac{h}{{2\pi }}} \right)\)
• Where n = 1, 2, 3, ..... each value of n corresponds to a permitted value of the orbit radius, r_n = Radius of nth orbit, v_n = corresponding speed.
• The radiation of energy occurs only when an electron jumps from one permitted orbit to another.

EXPLANATION:

CONCEPT:

• Bohr model: In 1913, Niels Bohr gave the Bohr's atom model which retained essential features of Rutherford's model and at the same time took into account its drawbacks.
• The electrons revolve around the nucleus in circular orbits which is called the stationary orbit.
• According to the Bohr's atom model, the electrons of an atom revolve around the nucleus only in those orbits in which the angular momentum of the electron is an integral multiple of h /2π.​
• By absorbing energy the electrons are able to jump from lower energy(E_i) level to higher energy level (E_f) and vice versa.
• The energy of emitted radiation is given by 

\(h\upsilon =E_F -E_i\)

• The energy of electrons in any orbit is given by:

\(E_n=~-13.6~\frac{{{Z}^{2}}}{{{n}^{2}}}\,eV\)

Where n is principal quantum number and Z is the atomic number.

CALCULATION:

Given - Ground state energy = -13.6 eV, n =2 (Since first excited state) Z =2

• The energy of the n^th excited state can be calculated using the equation 

\(\Rightarrow E_n = -13.6 \frac{Z^{2}}{n^{2}}\,eV\)

\(\Rightarrow E_n = -13.6 \frac{2^{2}}{2^{2}} = -13.6 eV\)

• The Energy of the first excited state will be \(E_n = -13.6 eV\)