Practice Start Test

Dual Nature of Radiation and Matter

NEET Physics · 96 questions · Page 3 of 10 · Click an option or "Show Solution" to reveal answer

Q21
An electron of mass m with an initial velocity v→=v0i^\overrightarrow v = {v_0}\widehat i (v 0 > 0) enters an electric field E→=−E→0i^\overrightarrow E = - {\overrightarrow E _0}\widehat i (E 0 = constant > 0) at t = 0. If λ\lambda 0 is its de-Broglie wavelength initially, then its de- Broglie wavelength at time t is
A λ0(1+eE0mv0t){{{\lambda _0}} \over {\left( {1 + {{e{E_0}} \over {m{v_0}}}t} \right)}}
B λ0(1+eE0mv0t){{\lambda _0}\left( {1 + {{e{E_0}} \over {m{v_0}}}t} \right)}
C λ\lambda 0 t
D λ\lambda 0
Correct Answer
Option A
Solution

Given, Initial velocity v = v 0

i^\widehat i

Electric field E = – E 0

i^\widehat i

∴\thereforeInitial de-Brogile wavelength, λ\lambda 0 =

hmv0{h \over {m{v_0}}}

Now force due to electric field on electrons,

F→=(−e)(−E0i^)=eE0i^\overrightarrow F = \left( { - e} \right)\left( { - {E_0}\widehat i} \right) = e{E_0}\widehat i

Acceleration produced in the electron,

a→=F→m=eE0mi^\overrightarrow a = {{\overrightarrow F } \over m} = {{e{E_0}} \over m}\widehat i

Now, velocity of electron after time t,

vt→=v→+a→t\overrightarrow {{v_t}} = \overrightarrow v + \overrightarrow a t

=

(v0+eE0tm)i^\left( {{v_0} + {{e{E_0}t} \over m}} \right)\widehat i

⇒\Rightarrow

∣vt→∣=v0+eE0tm\left| {\overrightarrow {{v_t}} } \right| = {v_0} + {{e{E_0}t} \over m}

Now, λ\lambda t =

hmvt{h \over {m{v_t}}}

=

hm(v0+eE0tm){h \over {m\left( {{v_0} + {{e{E_0}t} \over m}} \right)}}

=

hmv0(1+eE0tmv0){h \over {m{v_0}\left( {1 + {{e{E_0}t} \over {m{v_0}}}} \right)}}

=

λ0(1+eE0tmv0){{{\lambda _0}} \over {\left( {1 + {{e{E_0}t} \over {m{v_0}}}} \right)}}

[As

λ0=hmv0{\lambda _0} = {h \over {m{v_0}}}

]

Q22
When the light of frequency 2υ0{\upsilon _0} (where υ0{\upsilon _0} is threshold frequency), is incident on a metal plate, the maximum velocity of electrons emitted is v 1 . When the frequency of the incident radiation is increased to 5υ0{\upsilon _0} , the maximum velocity of electrons emitted from the same plate is v 2 . The ratio of v 1 to v 2 is
A 1 : 2
B 1 : 4
C 4 : 1
D 2 : 1
Correct Answer
Option A
Solution

From Einstein’s equation

12mv12=2hν0−hν0=hν0{1 \over 2}mv_1^2 = 2h{\nu _0} - h{\nu _0} = h{\nu _0}

Again,

12mv22=5hν0−hν0=4hν0{1 \over 2}mv_2^2 = 5h{\nu _0} - h{\nu _0} = 4h{\nu _0}

∴\therefore

v1v2=14=12{{{v_1}} \over {{v_2}}} = \sqrt {{1 \over 4}} = {1 \over 2}
Q23
The de-Broglie wavelength of a neutron in thermal equilibrium with heavy water at a temperature T (kelvin) and mass m, is
A h3mkT{h \over {\sqrt {3mkT} }}
B 2h3mkT{{2h} \over {\sqrt 3 mkT}}
C 2hmkT{{2h} \over {\sqrt {mkT} }}
D hmkT{h \over {\sqrt {mkT} }}
Correct Answer
Option A
Solution

Wavelength, λ\lambda =

hp{h \over p}

⇒\Rightarrow λ\lambda =

h2mE{h \over {\sqrt {2mE} }}

⇒\Rightarrow λ\lambda =

h2m×32kT{h \over {\sqrt {2m \times {3 \over 2}kT} }}

⇒\Rightarrow λ\lambda =

h3mkT{h \over {\sqrt {3mkT} }}
Q24
Electrons of mass m with de-Broglie wavelength λ\lambda fall on the target in an X-ray tube. The cutoff wavelength (λ\lambda 0 ) of the emitted X-ray is
A λ\lambda 0 = 2mcλ2h{{2mc{\lambda ^2}} \over h}
B λ0=2hmc{\lambda _0} = {{2h} \over {mc}}
C λ0=2m2c2λ3h2{\lambda _0} = {{2{m^2}{c^2}{\lambda ^3}} \over {{h^2}}}
D λ0=λ{\lambda _0} = \lambda
Correct Answer
Option A
Solution

Kinetic energy of electrons K =

p22m{{{p^2}} \over {2m}}

=

(hλ)22m{{{{\left( {{h \over \lambda }} \right)}^2}} \over {2m}}

=

h22mλ2{{{h^2}} \over {2m{\lambda ^2}}}

For certain frequency, maximum wavelength that can be emitted is λ\lambda 0 which is cut off wavelength obtained at cut off frequency, E 0 =

hcλ0{{hc} \over {{\lambda _0}}}

Since, E = E 0

h22mλ2=hcλ0{{{h^2}} \over {2m{\lambda ^2}}} = {{hc} \over {{\lambda _0}}}

⇒\Rightarrow λ\lambda 0 =

2mcλ2h{{2mc{\lambda ^2}} \over h}
Q25
Photons with energy 5 eV are incifent on a cathode C in a photoelectric cell. The maximum energy of emitted photoelectrons is 2 eV. When photons of energy 6 eV are incident on C, no photoelectrons will reach the anode A, if the stopping potential of A relative to C is
A + 3 V
B +4 V
C −-1 V
D −-3 V
Correct Answer
Option D
Solution

Stopping potential is the voltage which is needed to stop energetic photo electron for reaching towards cathode.

Stopping potential, ϕ\phi = E – K max 2 eV = 5 eV – ϕ\phi or ϕ\phi = 3 eV Now eV 0 = E' – ϕ\phi = 6 eV – 3 eV = 3 eV Hence stopping potential is –3V

Q26
An electron of mass m and a photon have same energy E. The ratio of de-Broglie wavelengths associated with them is
A c(2mE)12c{\left( {2mE} \right)^{{1 \over 2}}}
B 1c(2mE)12{1 \over c}{\left( {{{2m} \over E}} \right)^{{1 \over 2}}}
C 1c(E2m)12{1 \over c}{\left( {{E \over {2m}}} \right)^{{1 \over 2}}}
D (E2m)12{\left( {{E \over {2m}}} \right)^{{1 \over 2}}}
Correct Answer
Option C
Solution

For electron De-Broglie wavelength, λ\lambda e =

h2mE{h \over {\sqrt {2mE} }}

For photon of energy, E = hν\nu =

hcλp{{hc} \over {{\lambda _p}}}

⇒\Rightarrow λ\lambda p =

hcE{{hc} \over E}

∴\therefore

λeλp{{{\lambda _e}} \over {{\lambda _p}}}

=

h2mE×Ehc{h \over {\sqrt {2mE} }} \times {E \over {hc}}

=

1c(E2m)12{1 \over c}{\left( {{E \over {2m}}} \right)^{{1 \over 2}}}
Q27
When a metallic surface is illuminated with radiation of wavelength λ\lambda , the stopping potential is V. If the same surface is illuminated with radiation of wavelength 2 λ\lambda , the stopping potential is V4{V \over 4}. The threshold wavelength for the metallic surface is
A 52λ{5 \over 2}\lambda
B 3λ\lambda
C 4λ\lambda
D 5λ\lambda
Correct Answer
Option B
Solution

According to Einstein's photoelectric effect, eV =

hcλ{{hc} \over \lambda }

-

hcλ0{{hc} \over {{\lambda _0}}}

.......(1) e

V4{V \over 4}

=

hc2λ{{hc} \over 2\lambda }

-

hcλ0{{hc} \over {{\lambda _0}}}

.......(2) Dividing equation (i) by (ii) by, 4 =

1λ−1λ012λ−1λ0{{{1 \over \lambda } - {1 \over {{\lambda _0}}}} \over {{1 \over {2\lambda }} - {1 \over {{\lambda _0}}}}}

⇒\Rightarrow λ\lambda 0 = 3λ\lambda

Q28
A photoelectric surface is illuminated successively by monochromatic light of wavelength λ\lambda and λ/2\lambda/2 . If the maximum kinetic energy of the emitted photoelectrons in the second case is 3 times that in the first case, the work function of the surface of the material is (h = Planck's constant, C = speed of light)
A 2hcλ{{2hc} \over \lambda }
B hc3λ{{hc} \over {3\lambda }}
C hc2λ{{hc} \over {2\lambda }}
D hcλ{{hc} \over \lambda }
Correct Answer
Option C
Solution

According to Einstein’s photoelectric equation, the maximum kinetic energy of the emitted photoelectrons in the first case is

Kmax⁡1{K_{{{\max }_1}}}

=

hcλ{{hc} \over \lambda }

- ϕ\phi 0 and that in the second case is

Kmax⁡2{K_{{{\max }_2}}}

=

hcλ2{{hc} \over {{\lambda \over 2}}}

- ϕ\phi 0 Given

Kmax⁡2{K_{{{\max }_2}}}

= 3

Kmax⁡1{K_{{{\max }_1}}}

∴\therefore

2hcλ{{2hc} \over \lambda }

- ϕ\phi 0 = 3

(hcλ−ϕ0)\left( {{{hc} \over \lambda } - {\phi _0}} \right)

⇒\Rightarrow

2hcλ{{2hc} \over \lambda }

- ϕ\phi 0 =

3hcλ−3ϕ0{{{3hc} \over \lambda } - 3{\phi _0}}

⇒\Rightarrow 2ϕ\phi 0 =

hcλ{{{hc} \over \lambda }}

⇒\Rightarrow ϕ\phi 0 =

hc2λ{{hc} \over {2\lambda }}
Q29
Light of wavelength 500 nm is incifent on a metal with work function 2.28 eV. The de Broglie wavelength of the emitted electron is
A ≥ \ge 2.8 × \times 10 −-9 m
B ≤ \le 2.8 × \times 10 −-12 m
C < 2.8 × \times 10 −-10 m
D < 2.8 × \times 10 −-9 m
Correct Answer
Option A
Solution

KE max =

hcλ{{{hc} \over \lambda }}

- ϕ\phi ⇒\Rightarrow KE max =

6.6×10−34×3×1085×10−7×1.6×10−19{{6.6 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {5 \times {{10}^{ - 7}} \times 1.6 \times {{10}^{ - 19}}}}

- 2.28 ⇒\Rightarrow KE max = 2.48 – 2.28 = 0.2 eV λ\lambda min =

h2mKEmax⁡{h \over {\sqrt {2mK{E_{\max }}} }}

=

203×10342×9×10−31×0.2×1.6×10−19{{{{20} \over 3} \times {{10}^{34}}} \over {\sqrt {2 \times 9 \times {{10}^{ - 31}} \times 0.2 \times 1.6 \times {{10}^{ - 19}}} }}

= 2.8 ×\times 10 −-9 m ∴\therefore λ\lambda ≥\ge 2.8 ×\times 10 −-9 m

Q30
A certain metallic surface is illuminated with monochromatic light of wavelength, λ.\lambda . The stopping potential for photo-electric current for this light is 3V 0 . If the same surface is illuminated with light of wavelength 2 λ\lambda , the stopping potential is V 0 . The threshold wavelength for this surface for photo-electric effect is
A λ4{\lambda \over 4}
B λ6{\lambda \over 6}
C 6λ\lambda
D 4λ\lambda
Correct Answer
Option D
Solution

Stopping potential, ϕ\phi = E - K max 3eV 0 + ϕ\phi =

hcλ{{{hc} \over \lambda }}

......(1) eV 0 + ϕ\phi =

hc2λ{{{hc} \over 2\lambda }}

......(2) Solving equation (1) and (2), we get ϕ\phi =

hc4λ{{{hc} \over {4\lambda }}}

So, threshold wavelength, λ\lambda th =

hcϕ{{{hc} \over \phi }}

=

hchc4λ{{{hc} \over {{{hc} \over {4\lambda }}}}}

= 4λ\lambda

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