Practice Start Test

Dual Nature of Radiation and Matter

JEE Physics · 163 questions · Page 10 of 17 · Click an option or "Show Solution" to reveal answer

Q91
Assuming the nitrogen molecule is moving with r.m.s. velocity at 400 K, the de-Broglie wavelength of nitrogen molecule is close to : (Given : nitrogen molecule weight : 4.64 × \times 10–26 kg, Boltzman constant: 1.38 × \times 10–23 J/K, Planck constant : 6.63 × \times 10–34 J.s)
A 0.44 Ao\mathop A\limits^o
B 0.34 Ao\mathop A\limits^o
C 0.20 Ao\mathop A\limits^o
D 0.24 Ao\mathop A\limits^o
Correct Answer
Option D
Solution

We know, the de-Broglie wavelength λ\lambda =

hmvrms{h \over {m{v_{rms}}}}

also Vrms =

3kTm\sqrt {{{3kT} \over m}}

∴\therefore λ\lambda =

h3mkT{h \over {\sqrt {3mkT} }}

=

6.63×10−343×4.6×10−26×1.38×10−23×400{{6.63 \times {{10}^{ - 34}}} \over {\sqrt {3 \times 4.6 \times {{10}^{ - 26}} \times 1.38 \times {{10}^{ - 23}} \times 400} }}

= 2.4 ×\times 10-11 m = 0.24

Ao\mathop A\limits^o
Q92
Two electrons are moving with non-relativistic speed perpendicular to each other. If corresponding de Broglie wavelength are λ1{\lambda _1} and λ2,{\lambda _2}, their de Broglie wavelength in the frame of reference attached to their center of masses :
A λCM=λ1=λ2{\lambda _{CM}} = {\lambda _1} = {\lambda _2}
B λCM=2λ1λ2λ12+λ22{\lambda _{CM}} = {{2{\lambda _1}{\lambda _2}} \over {\sqrt {\lambda _1^2 + \lambda _2^2} }}
C 1λCM=1λ1+1λ2{1 \over {{\lambda _{CM}}}} = {1 \over {{\lambda _1}}} + {1 \over {{\lambda _2}}}
D λCM=(λ1+λ22){\lambda _{CM}} = \left( {{{{\lambda _1} + {\lambda _2}} \over 2}} \right)
Correct Answer
Option B
Solution

As we know, momentum (p) =

hλ{h \over \lambda }

Let one perticle is moving in x direction and other is in y dirrection.

∴    \therefore\,\,\,\,

momentum of each electrons

hλ1i^{h \over {{\lambda _1}}}\widehat i

and

hλ2j^{h \over {{\lambda _2}}}\widehat j
∴    \therefore\,\,\,\,

Velocity of each electrons

hmλ1i^{h \over {m{\lambda _1}}}\widehat i

and

hmλ2j^{h \over {m{\lambda _2}}}\widehat j
∴    \therefore\,\,\,\,

Velocity of center of mass

(υ→CM)\left( {{{\overrightarrow \upsilon }_{_{CM}}}} \right)

=

υ1i^+υ2j^2{{\upsilon _1^{\widehat i} + \upsilon _2^{\widehat j}} \over 2}

Now, velocity of first electron about center of mass,

υ→1CM{\overrightarrow \upsilon _{_{1CM}}}

=

υ1i^\upsilon _1^{\widehat i}

−-

(υ1i^+υ2j^2)\left( {{{\upsilon _1^{\widehat i} + \upsilon _2^{\widehat j}} \over 2}} \right)

=

υ1i^−υ2j^2{{\upsilon _1^{\widehat i} - \upsilon _2^{\widehat j}} \over 2}

Similarly,

υ→2CM=υ2j^−υ1i^2{\overrightarrow \upsilon _{_{2CM}}} = {{\upsilon _2^{\widehat j} - \upsilon _1^{\widehat i}} \over 2}

Here,

∣υ→1CM∣=∣υ→2CM∣=12\left| {{{\overrightarrow \upsilon }_{_{1CM}}}} \right| = \left| {{{\overrightarrow \upsilon }_{_{2CM}}}} \right| = {1 \over 2}
∴    \therefore\,\,\,\,
υ\upsilon

=

12{1 \over 2}
h2m2λ12+h2m2λ22\sqrt {{{{h^2}} \over {{m^2}\lambda _1^2}} + {{{h^2}} \over {{m^2}\lambda _2^2}}}

⇒\Rightarrow

    \,\,\,\,

m

υ\upsilon

=

12{1 \over 2}
h2λ12+h2λ22{\sqrt {{{{h^2}} \over {\lambda _1^2}} + {{{h^2}} \over {\lambda _2^2}}} }

⇒\Rightarrow

    \,\,\,\,
hλCM{h \over {{\lambda _{CM}}}}

= h

14λ12+14λ22{\sqrt {{1 \over {4\lambda _1^2}} + {1 \over {4\lambda _2^2}}} }

⇒\Rightarrow

    \,\,\,\,
1λCM=λ12+λ222λ1λ2{1 \over {{\lambda _{CM}}}} = {{\sqrt {\lambda _1^2 + \lambda _2^2} } \over {2{\lambda _1}{\lambda _2}}}

⇒\Rightarrow

    \,\,\,\,
λCM=2λ1λ2λ12+λ22{\lambda _{CM}} = {{2{\lambda _1}{\lambda _2}} \over {\sqrt {\lambda _1^2 + \lambda _2^2} }}
Q93
If the de Broglie wavelength of an electron is equal to the 10–3 times the wavelength of a photon of frequency 6 × \times 1014 Hz, then the speed of electron is equal to : (Speed of light = 3 × \times 108 m/s, Planck's constant = 6.63 × \times 10–34 J.s, Mass of electron = 9.1 × \times 10–31 kg)
A 1.7 × \times 106 m/s
B 1.45 × \times 106 m/s
C 1.1 × \times 106 m/s
D 1.8 × \times 106 m/s
Correct Answer
Option B
Solution
hmv=10−3(3×1086×1014){h \over {mv}} = {10^{ - 3}}\left( {{{3 \times {{10}^8}} \over {6 \times {{10}^{14}}}}} \right)

v

=6.63×10−34×6×10149.1×10−31×3×105= {{6.63 \times {{10}^{ - 34}} \times 6 \times {{10}^{14}}} \over {9.1 \times {{10}^{ - 31}} \times 3 \times {{10}^5}}}

v

=1.45×106= 1.45 \times {10^6}

m/s

Q94
A 2 mW laser operates at wavelength of 500 nm. The number of photons that will be emitted per second is : [Given Planck's constant h = 6.6 × 10–34 Js, speed of light c = 3.0 × 108 m/s]
A 5 × 1015
B 1.5 × 1016
C 1 × 1016
D 2 × 1016
Correct Answer
Option A
Solution
2×10−3=hcλdndt2 \times {10^{ - 3}} = {{hc} \over \lambda }{{dn} \over {dt}}
dndt=2×10−3λhc{{dn} \over {dt}} = {{2 \times {{10}^{ - 3}}\lambda } \over {hc}}
=2×10−3×500×10−96.6×10−34×3×108= {{2 \times {{10}^{ - 3}} \times 500 \times {{10}^{ - 9}}} \over {6.6 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}
=10006.6×3×10=5×1015= {{1000} \over {6.6 \times 3}} \times 10 = 5 \times {10^{15}}
Q95
In a photoelectric effect experiment the threshold wavelength of the light is 380 nm. If the wavelentgh of incident light is 260 nm, the maximum kinetic energy of emitted electrons will be: Given E (in eV) = 1237/λ\lambda (in nm)
A 4.5 eV
B 15.1 eV
C 3.0 eV
D 1.5 eV
Correct Answer
Option D
Solution
Kmax⁡=hcλ−hcλo{K_{\max }} = {{hc} \over \lambda } - {{hc} \over {{\lambda _o}}}
⇒Kmax⁡=hc(λo−λλλo)\Rightarrow {K_{\max }} = hc\left( {{{{\lambda _o} - \lambda } \over {\lambda {\lambda _o}}}} \right)
⇒Kmax⁡=(1237)(380−260380×260)\Rightarrow {K_{\max }} = \left( {1237} \right)\left( {{{380 - 260} \over {380 \times 260}}} \right)

= 1.5 eV

Q96
A particle 'P' is formed due to a completely inelastic collision of particles 'x' and 'y' having de-Broglie wavelengths 'λ\lambda x' and 'λ\lambda y' respectively. If x and y were moving in opposite directions, then the de-Broglie wavelength of 'P' is :-
A λx−λy{\lambda _x} - {\lambda _y}
B λxλy∣λx−λy∣{{{\lambda _x}{\lambda _y}} \over {\left| {{\lambda _x} - {\lambda _y}} \right|}}
C λx+λy{\lambda _x} + {\lambda _y}
D λxλyλx+λy{{{\lambda _x}{\lambda _y}} \over {{\lambda _x} + {\lambda _y}}}
Correct Answer
Option B
Solution

Conservation of momentum

px→+py→=p→final\overrightarrow {{p_x}} + \overrightarrow {{p_y}} = {\overrightarrow p _{final}}

mxvx – myvy = (mx + my) V

hλx−hλy=hλ{h \over {{\lambda _x}}} - {h \over {{\lambda _y}}} = {h \over \lambda }
⇒λ=λxλy∣λx−λy∣\Rightarrow \lambda = {{{\lambda _x}{\lambda _y}} \over {\left| {{\lambda _x} - {\lambda _y}} \right|}}
Q97
A metal plate of area 1 × \times 10–4 m2 is illuminated by a radiation of intensity 16 mW/m2. The work function of the metal is 5 eV. The energy of the incident photons is 10 eV and only 10% of it produces photo electrons. The number of emitted photoelectrons per second and their maximum energy, respectively, will be
A 1014 and 10 eV
B 1012 and 5 eV
C 1011 and 5 eV
D 1010 and 5 eV
Correct Answer
Option C
Solution

Given that the area of the metal plate

A=1×10−4m2A = 1 \times 10^{-4} m^2

, and the intensity of the radiation

I=16mW/m2=16×10−3W/m2I = 16 mW/m^2 = 16 \times 10^{-3} W/m^2

. The power

PP

falling on the metal plate is given by the product of the intensity of the radiation and the area:

P=IA=(16×10−3W/m2)(1×10−4m2)=1.6×10−6WP = IA = (16 \times 10^{-3} W/m^2)(1 \times 10^{-4} m^2) = 1.6 \times 10^{-6} W

. The energy of each incident photon is given as 10 eV. Converting this to joules (since 1 eV =

1.6×10−19J1.6 \times 10^{-19} J

):

Ephoton=10eV=10×1.6×10−19J=1.6×10−18JE_{\text{photon}} = 10 eV = 10 \times 1.6 \times 10^{-19} J = 1.6 \times 10^{-18} J

. The number of incident photons per second

nphotonn_{\text{photon}}

is then given by the total power divided by the energy per photon:

nphoton=P/Ephoton=1.6×10−6W/1.6×10−18J=1012 photons/sn_{\text{photon}} = P / E_{\text{photon}} = 1.6 \times 10^{-6} W / 1.6 \times 10^{-18} J = 10^{12} \text{ photons/s}

. Given that only 10% of these photons actually produce photoelectrons, the number of photoelectrons produced per second

nelectronn_{\text{electron}}

is:

nelectron=0.1×nphoton=0.1×1012=1011 electrons/sn_{\text{electron}} = 0.1 \times n_{\text{photon}} = 0.1 \times 10^{12} = 10^{11} \text{ electrons/s}

.

The work function of the metal is given as 5 eV, which is the energy required to remove an electron from the metal.

Therefore, the maximum energy of the photoelectrons

EmaxE_{\text{max}}

is given by the energy of the incident photons minus the work function:

Emax=Ephoton−Work function=10eV−5eV=5eVE_{\text{max}} = E_{\text{photon}} - \text{Work function} = 10 eV - 5 eV = 5 eV

.

Q98
In a Frank-Hertz experiment, an electron of energy 5.6 eV passes through mercury vapour and emerges with an energy 0.7 eV. The minimum wavelength of photons emitted by mercury atoms is close to :
A 2020 nm
B 250 nm
C 1700 nm
D 220 nm
Correct Answer
Option B
Solution

The minimum wavelength of emitted photons is λ\lambda =

12405.6−0.7nm{{1240} \over {5.6 - 0.7}}nm

= 250 nm

Q99
An electron, a doubly ionized helium ion (He++) and a proton are having the same kinetic energy. The relation between their respective de-Broglie wavelengths λ\lambda e, λ\lambda He++ and λ\lambda p is :
A λ\lambda e > λ\lambda He++ > λ\lambda p
B λ\lambda e < λ\lambda p < λ\lambda He++
C λ\lambda e > λ\lambda p > λ\lambda He++
D λ\lambda e < λ\lambda He++ = λ\lambda p
Correct Answer
Option C
Solution

λ\lambda =

hP{h \over P}

=

h2m(KE){h \over {\sqrt {2m\left( {KE} \right)} }}

∴\therefore λ\lambda ∝\propto

1m{1 \over {\sqrt m }}

mHe++ > mp > me ∴\therefore λ\lambdae > λ\lambdap > λ\lambdaHe++

Q100
When photon of energy 4.0 eV strikes the surface of a metal A, the ejected photoelectrons have maximum kinetic energy TA eV end de-Broglie wavelength λA\lambda _A. The maximum kinetic energy of photoelectrons liberated from another metal B by photon of energy 4.50 eV is TB = (TA – 1.5) eV. If the de-Broglie wavelength of these photoelectrons λB\lambda _B = 2λA\lambda _A, then the work function of metal B is :
A 1.5eV
B 4eV
C 2eV
D 3eV
Correct Answer
Option B
Solution

We know, de-Broglie wavelength λ\lambda =

hp=h2mKe{h \over p} = {h \over {\sqrt {2m{K_e}} }}

∴\therefore

λ∝1Ke\lambda \propto {1 \over {\sqrt {{K_e}} }}

So

λAλB=TBTA{{{\lambda _A}} \over {{\lambda _B}}} = \sqrt {{{{T_B}} \over {{T_A}}}}

⇒\Rightarrow

(12)2{\left( {{1 \over 2}} \right)^2}

=

TA−1.5TA{{{{T_A} - 1.5} \over {{T_A}}}}

On solving TA = 2 eV ∴\therefore TB = TA - 1.5 = 0.5 eV Also TB = 4.5 - ϕ\phiB ⇒\Rightarrow 0.5 = 4.5 - ϕ\phiB ⇒\Rightarrow ϕ\phiB = 4 eV

Ready for a full JEE mock test? Timed · full syllabus · instant results
Take a Mock Test →