Q: The photoelectric work function of potassium is 2.3eV. If light having a wavelength of 2800A˚ falls on potassium, find (a) the kinetic energy in electron volts of the most energetic electrons ejected. (b) the stopping potential in volts.

Given, W=2.3eV,λ=2800A˚ ∴E( in eV )=λ( in A˚)12375=280012375=34.4eV (a) Kmax=E−W=(4.4−2.3)eV=2.1eV (b) Kmax=eV0 ∴2.1eV=eV0 or V0=2.1 V

Q: In the above instance, what will be the maximum velocity of the photolectrons? (Mass of the electron =9.11×10−31 kg )

KEmaxvmax=m2KEmax=21mvmax2=9.11×10−312×2.42×10−19=0.7289×106 ms−1

Q: Photoelectric threshold of silver is λ=3800A˚. Ultraviolet light of λ=2600A˚ is incident on silver surface. What will be the value of work function?

(a) Here, λ0=3800A˚ Work function ϕ W=hv0=λ0hc=3800×10−106.63×10−34×3×108=5.23×10−19 J(3.2eV)

Q: The de-Broglie wavelength associated with an electron moving with a speed of 5.4×106 m/s is λe and for a ball of mass 150 g travelling at 30 m/s is λb, then which of the following relations between λe and λb is true?

(b) We know that de-Broglie wavelength is given by λ=ph where, h is Planck's constant and p is momentum. For the electron, mpλλem′p′λbλb=9.11×10−31 kg,v=5.4×106 m/s=mv=9.11×10−31×5.4×106=4.92×10−24 kg−m/s=ph=4.92×10−246.63×10−34=0.135 nm=0.150 kg,v′=30 m/s=m′v′=0.150×30=4.50 kg−m/s=p′h=4.50 kg−m/s6.63×10−34 J−s=1.47×10−34 m<λe

Q: Calculate the de-Broglie wavelength of neutrons whose energy is 1eV. (Given mass of electrons 1.67×10−27 kg )

(a) We know that, λ=2mEh=2×1.67×10−27×1.6×10−106.6×10−84=0.2857×10−10=0.2857A˚

Q: A metal has a work function of 2.0eV. It is illuminated by monochromatic light of wavelength 500 nm. Calculate (a) the threshold wavelength, (b) the maximum energy of photoelectrons, (c) the stopping potential. (Given, Planck's constant, h=6.6×10−34 Js, charge on electron, e=1.6×10−19C and 1eV=1.6×10−19C).

Here, h=6.6×10−34Js,e=1.6×10−19C,1eV=1.6×10−19 Jϕ=2.0eV=2.0×1.6×10−19=3.2×10−19 Jλ=500 nm=500×10−9 m (a) If λ0 is the threshold wavelength, then ϕ=λ0hc or λ0=ϕhc=3.2×10−196.6×10−34×3×108=618.75×10−9 m=618.75 nm (b) The maximum energy of the photoelectrons, 21mvmax2=hv−ϕ=λhc−ϕ=500×10−96.6×10−34×3×108−3.2×10−19=3.96×10−19−3.2×10−19=0.76×10−19 J (c) The stopping potential is given by eVV0 or V0=21mvmax2=0.76×10−19=e0.76×10−19=1.6×10−190.76×10−19=0.475 V

Q: What is the frequency and energy of a photon of wavelength 6000A˚ ? (Given h=6.6×10−34Js,C=3×108 ms−1 )

Frequency v=λC=6000×10−103×108=5×1014 Hz and energy =hv=6.6×10−34×5×1014=3.3×10−19 J

Question 1

The photoelectric work function of potassium is 2.3eV. If light having a wavelength of 2800A˚ falls on potassium, find
(a) the kinetic energy in electron volts of the most energetic electrons ejected.
(b) the stopping potential in volts.

Accepted Answer

Given, W=2.3eV,λ=2800A˚
∴E( in eV )=λ( in A˚)12375​=280012375​=34.4eV
(a) Kmax​=E−W=(4.4−2.3)eV=2.1eV
(b) Kmax​=eV0​
∴2.1eV=eV0​ or V0​=2.1 V

Question 2

In the above instance, what will be the maximum kinetic energy of the emitted photoelectrons?

Accepted Answer

(c) Incident wavelength, λ=2600A˚
∴ incident frequency
vKEmax​hvKEmax​​=λc​=2600×10−103×108​ Hz=hv−ϕ0​=2600×10−106.63×10−34×3×108​=7.65×10−19=4.78eV=hv−ϕ0​=4.78eV−3.27eV=1.51eV(2.42×10−19 J)​

Question 3

In the above instance, what will be the maximum velocity of the photolectrons? (Mass of the electron =9.11×10−31 kg )

Accepted Answer

KEmax​vmax​=m2KEmax​​​​=21​mvmax2​=9.11×10−312×2.42×10−19​​=0.7289×106 ms−1​

Question 4

Photoelectric threshold of silver is λ=3800A˚. Ultraviolet light of λ=2600A˚ is incident on silver surface. What will be the value of work function?

Accepted Answer

(a) Here, λ0​=3800A˚
Work function ϕ
W=hv0​​=λ0​hc​=3800×10−106.63×10−34×3×108​=5.23×10−19 J(3.2eV)​

Question 5

The de-Broglie wavelength associated with an electron, accelerated through a potential difference of 100 volts is

Accepted Answer

(c) Accelerating potential V=100 V
The de-Broglie wavelength,
λ=ph​=V​1.227​ nm
λ=100​1.227​ nm=0.123 nm

Question 6

A particle is moving three times as fast as an electron. The ratio of the de-Broglie wavelength of the particle to that of electron is 1.813×10−4. The mass of particle is

Accepted Answer

(a) de-Broglie wavelength of a moving particle, having mass m and velocity v is
λ=ph​=mvh​
Mass =m=λvh​
For an electron, mass me​=λe​ve​h​ ∴ ve​v​=3 and λe​λ​=1.813×10−4
 Then, mass of particle m=me​(λλe​​)(vve​​)m=(9.11×10−31 kg)×(31​)×(1.8131​×10−4)∴m=1.675×10−17 kg​ Note The particle with this mass could be a proton or a neutron.

Question 7

The de-Broglie wavelength associated with an electron moving with a speed of 5.4×106 m/s is λe​ and for a ball of mass 150 g travelling at 30 m/s is λb​, then which of the following relations between λe​ and λb​ is true?

Accepted Answer

(b) We know that de-Broglie wavelength is given by
λ=ph​
where, h is Planck's constant and p is momentum. For the electron,
mpλλe​m′p′λb​λb​​=9.11×10−31 kg,v=5.4×106 m/s=mv=9.11×10−31×5.4×106=4.92×10−24 kg−m/s=ph​=4.92×10−246.63×10−34​=0.135 nm=0.150 kg,v′=30 m/s=m′v′=0.150×30=4.50 kg−m/s=p′h​=4.50 kg−m/s6.63×10−34 J−s​=1.47×10−34 m<λe​​

Question 8

Calculate the de-Broglie wavelength of neutrons whose energy is 1eV. (Given mass of electrons 1.67×10−27 kg )

Accepted Answer

(a) We know that,
λ=2mE​h​​=2×1.67×10−27×1.6×10−10​6.6×10−84​=0.2857×10−10=0.2857A˚​

Question 9

A metal has a work function of 2.0eV.
It is illuminated by monochromatic light of wavelength 500 nm. Calculate (a) the threshold wavelength, (b) the maximum energy of photoelectrons, (c) the stopping potential. (Given, Planck's constant, h=6.6×10−34 Js, charge on electron, e=1.6×10−19C and 1eV=1.6×10−19C).

Accepted Answer

Here, h=6.6×10−34Js,e=1.6×10−19C,1eV=1.6×10−19 Jϕ=2.0eV=2.0×1.6×10−19=3.2×10−19 Jλ=500 nm=500×10−9 m​ (a) If λ0​ is the threshold wavelength, then ϕ=λ0​hc​ or λ0​=ϕhc​=3.2×10−196.6×10−34×3×108​=618.75×10−9 m=618.75 nm​ (b) The maximum energy of the photoelectrons, 21​mvmax2​​=hv−ϕ=λhc​−ϕ=500×10−96.6×10−34×3×108​−3.2×10−19=3.96×10−19−3.2×10−19=0.76×10−19 J​ (c) The stopping potential is given by eVV0​ or V0​​=21​mvmax2​=0.76×10−19=e0.76×10−19​=1.6×10−190.76×10−19​=0.475 V​

Question 10

What is the frequency and energy of a photon of wavelength 6000A˚ ? (Given h=6.6×10−34Js,C=3×108 ms−1 )

Accepted Answer

Frequency v=λC​=6000×10−103×108​=5×1014 Hz and energy =hv=6.6×10−34×5×1014=3.3×10−19 J

Question 11

The work function of sodium is 2.3eV. Calculate the maximum wavelength for the light that will cause photoelectrons to be emitted from the sodium.

Accepted Answer

KEmax​=hν−ϕ0​=λhc​−ϕ0​
λhc​−ϕ0​>0
(Kinetic energy is always positive)
λ<ϕ0​ch​
λmax​=ϕ0​ch​=2.3×1.6×10−193×108×6.62×10−34​=5400A˚

Question 12

In an experiment on photoelectric effect it was observed that for incident light of wavelength 1.98×10−7 m, stopping potential is 2.5 V. What is the energy of photoelectrons with maximum speed, work function ϕ0​ and threshold frequency?

Accepted Answer

KEmax​=eV=2.5eV
Energy of the incident light
E​=λhc​=1.98×10−76.6×10−34×3×108​=1.98×10−7×1.6×10−196.6×10−34×3×108​eV=6.25eV​ Now, work function
(ϕ0​)=E−KE=3.75eV Moreover ϕ0​=hv0​⇒v0​=hϕ0​​∴v0​=6.6×10−343.75×1.6×10−19​=9.10×1014 Hz​

Question 13

Find the ratio of de-Broglie wavelength of proton and α-particle which has been accelerated through the same potential difference.

Accepted Answer

λ=2mqV​h​,λα​λp​​=2mp​qp​V2mα​qα​V​​
As, mα​mp​​=41​ and qp​qα​​=12​⇒λα​λp​​=4×2​=22​

Question 14

The anode voltage of a photocell is kept fixed. The wavelength λ of the light falling on the cathode is gradually changed. The plate current i of the photocell varies as follows.

Accepted Answer

On increasing wavelength of light the photoelectric current decreases and at a certain wavelength (cut-off) above which photoelectric current stops. Hence, graph (a) is correct.

Question 15

The ratio of de-Broglie wavelengths of molecules of hydrogen and helium which are at temperature 27∘C and 127∘C respectively is

Accepted Answer

Wavelength λ=mvrms ​h​=3mKT​h​
=λHe​λH​​=mH​TH​mHe​THe​​​=2(273+27)4(273+127)​​=38​​

Question 16

A particle is dropped from a height H. The de-Broglie wavelength of the particle as a function of height is proportional to

Accepted Answer

Velocity acquired by a particle while falling from a height H is, v=2gH​
λ=mvh​=m2gH​h​ or λ∝H−21​

Question 17

A parallel beam of light is incident normally on a plane surface absorbing 40% of the light and reflecting the rest. If the incident beam carries 60 W of power, the force exerted by it on the surface is

Accepted Answer

Momentum of incident light per second
p1​=cE​=3×10860​=2×10−7
Momentum of reflected light per second
p2​=10060​×cE​=3×10860​=1.2×10−7
∴ Force on the surface = change in momentum per second
=p2​−(−p1​)=p2​+p1​=(2+1.2)×10−7=3.2×10−7 N​

Question 18

Calculate the energy of a photon with momentum 3.3×10−13 kg−ms−1, given Planck's constant to be 6.6×10−34Js

Accepted Answer

E/CE​=3.3×10−13 kg ms−1=3.3×10−13×c=3.3×10−13×3×108=9.9×10−5 J​

Question 19

A particle of mass 1mg has the same wavelength as an electron moving with a velocity of 3×106 ms−1. The velocity of the particle is

Accepted Answer

de-Broglie wavelength, λ=mvh​
As both particle and electrons having same wavelength therefore, their momentum will be equal to
mp​vp​vp​vp​​=me​ve​=mp​me​ve​​=10−69.1×10−31×3×106​=2.7×10−18 m/s​

Question 20

The energy that should be added to an electron to reduce its de-Broglie wavelength from 10−10 m to 0.5×10−10 m, will be

Accepted Answer

λλλ2​λ1​​0.5×10−1010−10​E2​​=2mE​h​∝E​1​=E1​E2​​​=E1​E2​​​=4E1​​
Therefore,
added energy =E2​−E1​=3E1​

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Solutions for Master Resource Book in JEE Main Physics by D. B. Singh

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