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Structure of Atom

JEE Chemistry · 129 questions · Page 8 of 13 · Click an option or "Show Solution" to reveal answer

Q71
For any given series of spectral lines of atomic hydrogen, let Δv_=\Delta \mathop v\limits^\_ = Δv‾max⁡−Δv‾min⁡\Delta {\overline v _{\max }} - \Delta {\overline v _{\min }} be the difference in maximum and minimum frequencies in cm–1. The ratio Lyman Balmer Δv‾LymanΔv‾Balmer{{\Delta {{\overline v }_{Lyman}}} \over {\Delta {{\overline v }_{Balmer}}}} is :
A 9 : 4
B 4 : 1
C 27 : 5
D 5 : 4
Correct Answer
Option A
Solution

We know,

v‾=RZ2(1n12−1n22)\overline v = R{Z^2}\left( {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right)

∴\therefore For Lyman series,

Δv‾Lyman{\Delta {{\overline v }_{Lyman}}}

=

Δv‾max⁡−Δv‾min⁡{\Delta {{\overline v }_{\max }} - \Delta {{\overline v }_{\min }}}

=

[11−1∞]−[11−14]\left[ {{1 \over 1} - {1 \over \infty }} \right] - \left[ {{1 \over 1} - {1 \over 4}} \right]

=

14{{1 \over 4}}

∴\therefore For Balmer series,

Δv‾Balmer{\Delta {{\overline v }_{Balmer}}}

=

Δv‾max⁡−Δv‾min⁡{\Delta {{\overline v }_{\max }} - \Delta {{\overline v }_{\min }}}

=

[14−1∞]−[14−19]\left[ {{1 \over 4} - {1 \over \infty }} \right] - \left[ {{1 \over 4} - {1 \over 9}} \right]

=

19{{1 \over 9}}

∴\therefore

Δv‾LymanΔv‾Balmer{{\Delta {{\overline v }_{Lyman}}} \over {\Delta {{\overline v }_{Balmer}}}}

=

1419{{{1 \over 4}} \over {{1 \over 9}}}

=

94{{9 \over 4}}
Q72
Energy of an electron is given by E=−2.178×10−18J(Z2n2)E = - 2.178 \times {10^{ - 18}}J\left( {{{{Z^2}} \over {{n^2}}}} \right). Wavelength of light required to excite an electron in an hydrogen atom from level n = 1 to n = 2 will be (h = 6.62 × 10−34 Js and c = 3.0 × 108 ms−1)
A 2.816 × 10−7 m
B 6.500 × 10−7 m
C 8.500 × 10−7 m
D 1.214 × 10−7 m
Correct Answer
Option D
Solution

Given

EE

=

−2.178×10−18J(Z2n2)- 2.178 \times {10^{ - 18}}J\left( {{{{Z^2}} \over {{n^2}}}} \right)

=

−2.178×10−8Z2(1n22−1n12)- 2.178 \times {10^{ - 8}}{Z^2}\left( {{1 \over {n_2^2}} - {1 \over {n_1^2}}} \right)

Electron in an hydrogen atom exited from level n = 1 to n = 2. So n2 = 2 and n1 = 1. And for H, Z = 1 ∴\therefore

E=−2.178×10−8(1(2)2−1(1)2)E = - 2.178 \times {10^{ - 8}}\left( {{1 \over {{{\left( 2 \right)}^2}}} - {1 \over {{{\left( 1 \right)}^2}}}} \right)

= 1.6335 ×\times 10-18 J We know

EE

=

hcλ{{hc} \over \lambda }

∴\therefore

hcλ{{hc} \over \lambda }

= 1.6335 ×\times 10-18 then

λ=6.62×10−34×3×1081.6335×10−18\lambda = {{6.62 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {1.6335 \times {{10}^{ - 18}}}}
=1.214×10−7m= 1.214 \times {10^{ - 7}}m
Q73
A stream of electrons from a heated filament was passed between two charged plates kept at a potential difference V esu. If e and m are charge and mass of an electron, respectively, then the value of h / λ\lambda (where λ\lambda is wavelength associated with electron wave) is given by :
A 2meV
B meV\sqrt {meV}
C 2meV\sqrt {2meV}
D meV
Correct Answer
Option C
Solution

We know, Kinetic Energy (KE) =

12mv2{1 \over 2}m{v^2}

∴\therefore mv2 = 2KE ⇒\Rightarrow m2v2 = 2mKE ⇒\Rightarrow mv =

2mKE\sqrt {2m{K_E}}

For an electron of charge 'e' which is passes through 'V' volt, kinetic energy of electron will be KE = eV ∴\therefore m =

2meV\sqrt {2meV}

We know de-Broglie wavelength for an electron

(λ)\left( \lambda \right)

=

hp=hmv{h \over p} = {h \over {mv}}

∴\therefore

hλ{h \over \lambda }

= mv =

2meV\sqrt {2meV}
Q74
What is the work function of the metal if the light of wavelength 4000A∘\mathop A\limits^ \circ generates photoelectrons of velocity 6 × \times 105 ms–1 from it ? (Mass of electron = 9 × \times 10–31 kg; Velocity of light = 3 × \times 108 ms−-1 Plank's constant = 6.626 × \times 10–34 Js; Charge of electron = 1.6 × \times 10–19 JeV–1)
A 4.0 eV
B 0.9 eV
C 2.1 eV
D 3.1 eV
Correct Answer
Option C
Solution

E = ϕ\phi + K.E ⇒\Rightarrow hν\nu = ϕ\phi +

12mv2{1 \over 2}m{v^2}

⇒\Rightarrow ϕ\phi =

hν−12mv2h\nu - {1 \over 2}m{v^2}

=

6.626×10−34×3×1084000×10−10−12×9×10−31×(6×105)2{{6.626 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {4000 \times {{10}^{ - 10}}}} - {1 \over 2} \times 9 \times {10^{ - 31}} \times {\left( {6 \times {{10}^5}} \right)^2}

= 3.35 ×\times 10-19 J ⇒\Rightarrow ϕ\phi =

3.35×10−191.6×10−19{{3.35 \times {{10}^{ - 19}}} \over {1.6 \times {{10}^{ - 19}}}}

eV = 2.0934 eV

≃\simeq

2.1 eV

Q75
The region in the electromagnetic spectrum where the Balmar series lines appear is :
A Microwave
B Ultraviolet
C Visible
D Infrared
Correct Answer
Option C
Solution

In the hydrogen spectrum, Balmer series lies in visible region.

Q76
If the de Broglie wavelength of the electron in nth Bohr orbit in a hydrogenic atom is equal to 1.5 π\pi a0 (a0 is Bohr radius), then the value of n/z is -
A 0.75
B 0.40
C 1.50
D 1.0
Correct Answer
Option A
Solution

According to debroglie hypothesis, 2π\pirn = nλ\lambda ∴\therefore λ\lambda =

2πrnn{{2\pi {r_n}} \over n}

According to the question,

2πrnn{{2\pi {r_n}} \over n}

= 1.5 π\pia0 We know, rn = 0.529

n2Z{{{n^2}} \over Z}

= a0 ×\times

n2Z{{{n^2}} \over Z}

∴\therefore

2πn×a0n2Z{{2\pi } \over n} \times {{{a_0}{n^2}} \over Z}

= 1.5 π\pi a0 ⇒\Rightarrow

nZ{n \over Z}

= 0.75

Q77
Uncertainty in the position of an electron (mass = 9.1 ×\times 10-31 kg) moving with a velocity 300 ms-1, accurate upto 0.001% will be (h = 6.63 ×\times 10-34 Js)
A 1.92 ×\times 10-2 m
B 3.84 ×\times 10-2 m
C 19.2 ×\times 10-2 m
D 5.76 ×\times 10-2 m
Correct Answer
Option A
Solution

% error in velocity =

ΔVV×100{{\Delta V} \over V} \times 100

∴\therefore 0.001 =

ΔV300×100{{\Delta V} \over {300}} \times 100

⇒\Rightarrow

Δ\Delta

V = 3 ×\times 10-3 According to Heisenberg uncertainty principle,

Δx.mΔV≥h4π\Delta x.m\Delta V \ge {h \over {4\pi }}

⇒\Rightarrow

Δx=h4πmΔV\Delta x = {h \over {4\pi m\Delta V}}

⇒\Rightarrow

Δx=6.63×10−344×3.14×9.1×10−31×3×10−3\Delta x = {{6.63 \times {{10}^{ - 34}}} \over {4 \times 3.14 \times 9.1 \times {{10}^{ - 31}} \times 3 \times {{10}^{ - 3}}}}

= 1.92 ×\times 10-2 m

Q78
In an atom, an electron is moving with a speed of 600 m/s with an accuracy of 0.005%. Certainity with which the position of the electron can be located is (h = 6.6 ×\times 10-34 kg m2s-1, mass of electron, em = 9.1 ×\times 10-31 kg)
A 5.10 ×\times 10-3 m
B 1.92 ×\times 10-3 m
C 3.84 ×\times 10-3 m
D 1.52 ×\times 10-4 m
Correct Answer
Option B
Solution

% error in velocity =

ΔVV×100{{\Delta V} \over V} \times 100

∴\therefore 0.005 =

ΔV600×100{{\Delta V} \over {600}} \times 100

⇒\Rightarrow

Δ\Delta

V = 3 ×\times 10-2 According to Heisenberg uncertainty principle,

Δx.mΔV≥h4π\Delta x.m\Delta V \ge {h \over {4\pi }}

⇒\Rightarrow

Δx=h4πmΔV\Delta x = {h \over {4\pi m\Delta V}}

⇒\Rightarrow

Δx=6.63×10−344×3.14×9.1×10−31×3×10−2\Delta x = {{6.63 \times {{10}^{ - 34}}} \over {4 \times 3.14 \times 9.1 \times {{10}^{ - 31}} \times 3 \times {{10}^{ - 2}}}}

= 1.92 ×\times 10-3 m

Q79
The energy of one mole of photons of radiation of wavelength 300 nm is (Given : h = 6.63 ×\times 10−-34 J s, NA = 6.02 ×\times 1023 mol−-1, c = 3 ×\times 108 m s−-1)
A 235 kJ mol−-1
B 325 kJ mol−-1
C 399 kJ mol−-1
D 435 kJ mol−-1
Correct Answer
Option C
Solution

Energy of one photon

E=1240λ(nm)eVE = {{1240} \over {\lambda (nm)}}eV
=1240300= {{1240} \over {300}}
=4.1333 eV= 4.1333\,eV

∴\therefore Energy of one mole of photon

=4.1333×6.02×1023 eV= 4.1333 \times 6.02 \times {10^{23}}\,eV
=4.1333×6.02×1023×1.6×10−19 J= 4.1333 \times 6.02 \times {10^{23}} \times 1.6 \times {10^{ - 19}}\,J
=4.1333×6.02×1023×1.6×10−191000 kJ= {{4.1333 \times 6.02 \times {{10}^{23}} \times 1.6 \times {{10}^{ - 19}}} \over {1000}}\,kJ
=399 kJ/mol= 399\,kJ/mol
Q80
If aa0 is denoted as the Bohr radius of hydrogen atom, then what is the de-Broglie wavelength (λ) of the electron present in the second orbit of hydrogen atom? [n : any integer]
A 8πa0n\dfrac{8 \pi a_0}{n}
B 2a0nπ\dfrac{2 a_0}{n \pi}
C na04π\dfrac{n a_0}{4 \pi}
D 4πa0n\dfrac{4 \pi a_0}{n}
Correct Answer
Option A
Solution

Bohr radius of hychogon atom →a0\rightarrow a_0 According to Bohr, the equation used to calculate the angular momentum of an election in a hydrogen atom is mvr=nh2π\mathrm{mvr=\dfrac{nh}{2\pi}} ..... (1) m→m \rightarrow mass of electron v→v \rightarrow velocity of electron r→r \rightarrow radius of the orbit n→n \rightarrow orbit. number. in. which electron is present.

Given that; election is present in second orbit, n=2n=2 The radius of the second orbit r2=a0×22=4a0r_2=a_0\times2^2=4a_0 General formula for radius of nthn^{th} orbit, rn=a0×n2r_n=a_0\times n^2 From (1)

mvr=nh2π2πr=nhmv\begin{aligned} & m v r=n \frac{h}{2 \pi} \\ & 2 \pi r=n \frac{h}{m v} \end{aligned}

hmv=λ\dfrac{h}{m v}=\lambda (de Broglie relation ship, λ→\lambda \rightarrow de Broglie Wavelength So, 2πr=nλ2 \pi r=n \lambda For the electron in the second orbit, 2πr2=nλ2 \pi r_2=n \lambda Substitute for r2r_2,

2π×4a0=nλ8πa0=nλ∴λ=8πa0n\begin{aligned} & 2 \pi \times 4 a_0=n \lambda \\ & 8 \pi a_0=n \lambda \\ & \therefore \lambda=\frac{8 \pi a_0}{n} \end{aligned}

Correct answer: Option 1) 8πa0n\dfrac{8 \pi a_0}{n}

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