Q: Photoelectrons are ejected from a metal surface using photons of energy 4×10−20 J. The de Broglie wavelength of the electron emitted with maximum K.E. is 59A∘ . What is the photoelectric threshold in joules?

E≡ϕ+(K⋅E)max Where E is the energy incident, ϕ is the work function.4×10−20=ϕ+2mp2where p is the momentum and m is the mass of electron. From de Broglie's equation, we have p=λh4×10−20=ϕ+2×(9.1×10−31)(59×10−10)2(6.62×10−34)2⇒ϕ=3.31×10−20 J.

Q: A 50 watt bulb emits monochromatic red light of wavelength of 795 nm . The number of photons emitted per second by the bulb is x×1020 . The value of x is ____ (Nearest integer)[Given : h=6.63×10−34Js and c=3.0×108 ms−1 ]

Power =120 J/sec .E=n[λhc]120=n[6.62×10−34×3×108120×920×10−9]=5558.9×1017=5.5589×1020

Q: A source of monochromatic radiation of wavelength 400 nm provides 1000 J of energy in 10 seconds. When this radiation falls on the surface of sodium, x×1020 electrons are ejected per second. Assume that wavelength 400 nm is sufficient for ejection of electron from the surface of sodium metal. The value of x is. (Nearest integer)(h=6.626×10−34Js)

Energy per second =100 J E=E0+KE so E=E0=[λhc]={400×10−96.62×10−32×3×108}=0.04965×10−17 J=4.965×10−19 J Number of of electron ejected by 100 J energy =4.965×10−19100=20.140×1019=2.01×1020

Q: The number of photons emitted by a monochromatic (single frequency) infrared range finder of power 1 mW and wavelength of 1000 nm, in 0.1 second is x×1013. The value of x is (Nearest integer) (h=6.63×10−34Js,c=3.00×108 ms−1)

Energy emitted in 0.1 sec=0.1sec⋅×10−3sJ=10−4 J If 'n' photons of λ=1000 nm are emitted, then 10−4=n=λhc⇒10−4=1000×10−9n×6.63×10−34×3×108 ⇒n=5.02×1014=50.2×1013 ⇒50 (nearest integer)

Q: A metal surface is exposed to 500 nm radiation. The threshold frequency of the metal for photoelectric current is 4.3×1014 Hz. The velocity of ejected electron is ×105 ms−1. (Nearest integer) [Use: h=6.63×10−34Js,me=9.0×10−31]

λ=500 nm→v=λC=500×10−93×108=6×1014 Hzv0=4.3×1014 Hz For PEE, hv =hv0+KεKE=hv−hv0=h(v−v0)=6.6×10−34(6×1014−4.3×1014)=6.6×1.7×10−20 J K.E.=21mv2;v=m2×K.E.V=9.1×10−312×6.6×1.7×10−20=24.65×1010=5×105 m/s

Q: Which of the graphs shown below does not represent the relationship between incident light and the electron ejected from metal surface?

For photoelectric effect,(i) KE=E−E0where,KE= Kinetic energy of ejected electrons.E= Energy of incident light =hvE0= Threshold energy =hv0v= Frequency of incident lightv0= Threshold frequencySlope =±1, intercept =−E0So, option (a) is correct.(ii) KE of ejected electrons does not depend on the intensity of incident light So, option (b) is correct. (iii) When, number of ejected electrons is plotted with frequency of light, we getSo, option (c) is also correct.(iv) KE=hv−hv0Slope =+h, intercept =−hv0So, option (d) is not correct.

Q: 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 (where, λ is wavelength associated with electron wave) is given by (a) 2meV (b) meV (c) 2meV (d) meV

De-Broglie wavelength for an electron (λ)=ph ⇒p=λh ..(i) Kinetic energy of an electron =eV As we know that, KE=2mp2 ∴eV=2mp2 or p=2meV From equations (i) and (ii), we get λh=2meV

Q: When a certain metal was irradiated with light of frequency 3.2×1016 Hz, photoelectrons emitted had twice the kinetic energy as did photoelectrons emitted when the same metal was irradiated with light of frequency 2.0×1016 Hz. Hence, threshold frequency is

(b) hvKE(KE)1(KE)2(KE)1 Given, h(3.2×1016−v0)∴v0=hv0+(KE)=h(v−v0)=h(3.2×1016−v0)=h(2.0×1016−v0)=2(KE)2=2h(2.0×1016−v0)=8.0×1015 Hz

Q: Kinetic energy of an article of 20 g is 4 J while that of another particle of the same mass is 16 J. Hence, velocity ratio of their motion is

(c)KE=21mv2E2E1=v22v12v2v1=E2E1=1:2

Question 1

Photoelectrons are ejected from a metal surface using photons of energy 4×10−20 J. The de Broglie wavelength of the electron emitted with maximum K.E. is 59A∘ . What is the photoelectric threshold in joules?

Accepted Answer

E≡ϕ+(K⋅E)max​ Where E is the energy incident, ϕ is the work function.4×10−20=ϕ+2mp2​where p is the momentum and m is the mass of electron. From de Broglie's equation, we have p=λh​4×10−20=ϕ+2×(9.1×10−31)(59×10−10)2(6.62×10−34)2​⇒ϕ=3.31×10−20 J.

Question 2

A 50 watt bulb emits monochromatic red light of wavelength of 795 nm . The number of photons emitted per second by the bulb is x×1020 . The value of x is ____ (Nearest integer)[Given : h=6.63×10−34Js and c=3.0×108 ms−1 ]

Accepted Answer

Power =120 J/sec .​E=n[λhc​]120=n[6.62×10−34×3×108120×920×10−9​]=5558.9×1017=5.5589×1020​

Question 3

A source of monochromatic radiation of wavelength 400 nm provides 1000 J of energy in 10 seconds. When this radiation falls on the surface of sodium, x×1020 electrons are ejected per second. Assume that wavelength 400 nm is sufficient for ejection of electron from the surface of sodium metal. The value of x is. (Nearest integer)(h=6.626×10−34Js)

Accepted Answer

Energy per second =100 J E=E0​+KE so E=E0​=[λhc​]​={400×10−96.62×10−32×3×108​}=0.04965×10−17 J=4.965×10−19 J​ Number of of electron ejected by 100 J energy =4.965×10−19100​=20.140×1019=2.01×1020

Question 4

The number of photons emitted by a monochromatic (single frequency) infrared range finder of power 1 mW and wavelength of 1000 nm, in 0.1 second is x×1013. The value of x is (Nearest integer) (h=6.63×10−34Js,c=3.00×108 ms−1)

Accepted Answer

Energy emitted in 0.1 sec​=0.1sec⋅×10−3sJ​=10−4 J​ If 'n' photons of λ=1000 nm are emitted, then 10−4=n=λhc​⇒10−4=1000×10−9n×6.63×10−34×3×108​ ⇒n=5.02×1014=50.2×1013 ⇒50 (nearest integer)

Question 5

A metal surface is exposed to 500 nm radiation. The threshold frequency of the metal for photoelectric current is 4.3×1014 Hz. The velocity of ejected electron is ×105 ms−1. (Nearest integer) [Use: h=6.63×10−34Js,me​=9.0×10−31]

Accepted Answer

​λ=500 nm→v=λC​=500×10−93×108​=6×1014 Hzv0​=4.3×1014 Hz For PEE, hv =hv0​+KεKE=hv−hv0​=h(v−v0​)=6.6×10−34(6×1014−4.3×1014)=6.6×1.7×10−20 J K.E.=21​mv2;v=m2×K.E.​​V=9.1×10−312×6.6×1.7×10−20​​=24.65×1010​=5×105 m/s​

Question 6

Which of the graphs shown below does not represent the relationship between incident light and the electron ejected from metal surface?

Accepted Answer

For photoelectric effect,(i) KE=E−E0​where,KE= Kinetic energy of ejected electrons.E= Energy of incident light =hvE0​= Threshold energy =hv0​v= Frequency of incident lightv0​= Threshold frequencySlope =±1, intercept =−E0​So, option (a) is correct.(ii) KE of ejected electrons does not depend on the intensity of incident light

So, option (b) is correct.
(iii) When, number of ejected electrons is plotted with frequency of light, we getSo, option (c) is also correct.(iv) KE=hv−hv0​Slope =+h, intercept =−hv0​So, option (d) is not correct.

Question 7

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 (where, λ is wavelength associated with electron wave) is given by
(a) 2meV
(b) meV​
(c) 2meV​
(d) meV

Accepted Answer

De-Broglie wavelength for an electron (λ)=ph​

⇒p=λh​ ..(i)

Kinetic energy of an electron =eV

As we know that, KE=2mp2​

∴eV=2mp2​ or p=2meV​

From equations (i) and (ii), we get λh​=2meV​

Question 8

Threshold energy is also called

Accepted Answer

(a) work function

Question 9

When a certain metal was irradiated with light of frequency 3.2×1016 Hz, photoelectrons emitted had twice the kinetic energy as did photoelectrons emitted when the same metal was irradiated with light of frequency 2.0×1016 Hz. Hence, threshold frequency is

Accepted Answer

(b) hvKE(KE)1​(KE)2​(KE)1​ Given, h(3.2×1016−v0​)∴v0​​=hv0​+(KE)=h(v−v0​)=h(3.2×1016−v0​)=h(2.0×1016−v0​)=2(KE)2​=2h(2.0×1016−v0​)=8.0×1015 Hz​

Question 10

Kinetic energy of an article of 20 g is 4 J while that of another particle of the same mass is 16 J. Hence, velocity ratio of their motion is

Accepted Answer

(c)KE=21​mv2E2​E1​​=v22​v12​​v2​v1​​=E2​E1​​​=1:2

Question 11

A substance mass of 10 kg at height of 10 m is allowed to fall. A place where kinetic energy is 200 J, potential energy is

Accepted Answer

(b) PE=mgh=10×10×10=1000 Jat the height of 10 m.When body falls potential energy decreases and kinetic energy increases such thatKE+PE= total energy ∴PE=1000−200=800 J

Question 12

When electromagnetic radiation of wavelength 300 nm falls on the surface of sodium, electrons are emitted with a kinetic energy of 1.68×105 J mol−1. The minimum energy needed to remove an electron from sodium is ...........10−19?Answer in 3 sig.

Accepted Answer

hλc​=hvo​+K.E
hvo​=300×10−9m6.626×10−34Js×3×108ms−1​−6.022×1022mol−11.68×105Jmol−1​=3.84×10−19J

Question 13

For silver metal, threshold frequency for the emission of photoelectron is 1.13×107 s−1. The kinetic energy of electrons when silver is irradiated with a wavelength of 1.5 nm is:

Accepted Answer

We haveE=λhc​−hv0​=h(λc​−v0​)=(6.626×10−34 J s)[1.5×10−9 m3×108 m s−1​−1.13×107 s−1]=(6.266×10−34 J s)(0.87×1017 s−1)=5.76×10−17 J

Question 14

The binding energy of electrons in a metal is 193 kJ mol−1. The threshold frequency of the metal is:

Accepted Answer

We have NA​hv= Binding energy per molev=(6.022×1023 mol−1)(6.626×10−34 J s)(193×103 J mol−1)​=4.84×1014 s−1

Question 15

The work function of a substance is 4.0eV. The longest wavelength of light that can cause photoelectron emission from this substance is approximately

Accepted Answer

λmax​=Whc​=4(1.6×10−19 J)(6.626×10−34 J s)(3×108 ms−1)​=3.10×10−7 m=310 nm

Question 16

Photoelectric effect supports quantum nature of light because

Accepted Answer

(a) there is a minimum frequency of light below which no photoelectron is emitted.(c) even when the metal surface is faintly illuminated, the photoelectrons are emitted

Question 17

When photons of energy 4.25eV strike the surface of a metal A, the ejected photoelectrons have maximum kinetic energy and de-Broglie wave length is λA​. The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.70eV is (KE)B​=(KE)A​−1.50eV. If the de-Broglie wavelength of these photoelectrons is λB​=2λA​, then.

Accepted Answer

The ejected photoelectrons satisfy the expression Energy of photon = Work function + KE of electronHence, 4.25eV=(KE)A​+WA​ and 4.70eV=(KE)B​+WB​The expression of de Broglie wavelength is λ=ph​=2mE​h​( since E=21​mv2=2mp2​)HenceλB​λA​​=[(KE)A​(KE)B​​]1/2=21​ (given )⇒(KE)A​=4(KE)B​ Since  Thus ​(KE)B​=(KE)A​−1.50eV, we get (KE)B​=4(KE)B​−1.50eV or (KE)B​=0.50eV(KE)=200eV​Thus (KE)A​=2.00eV

Question 18

When the frequency of light incident on a metallic plate is doubled, the K.E. of the emitted photoelectron will be

Accepted Answer

(c)hv1​=hv0​+K⋅E1​
2hv1​=hv0​+K⋅E⋅1​
 K.E. =hv1​−hv0​
The value of K.E. will increase but more than double of the previous K.E.

Question 19

The work function for a metal is 4eV. To emit a photo electron of zero velocity from the surface of the metal, the wavelength of incident light should be (nearly)

Accepted Answer

Given: ϕmetal ​=4eV
 K.E. =0
Applying photoelectric equation,
K.E.=E−ϕ⇒E=ϕ​ 
We know, 
E(eV)=λA˚∘12400​⇒λ=E(eV)12400​=412400​=3100A˚​
So the required wavelength of incident light should be 3100A˚

Question 20

Electromagnetic radiations having λ=310A˚ are subjected to a metal sheet having work function =12.8 eV. What will be the velocity of photoelectrons with maximum kinetic energy.

Accepted Answer

(c)E=λhc​⇒λnm​1240​eVE=311240​⇒40eV40=12.8+K.E. K.E. =40−12.8=27.2eV K.E. =27.2×1.6×10−1927.2×1.6×10−19=21​×9.1×10−31×v2v=2.18×2​×106 m/s

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