QBManager

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357726 The Einstein's photoelectric equation is based upon the conservation of

1 Mass

2 Momentum

3 Angular momentum

4 Energy

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357727 In photoelectric experiment energy of $2.48 e V$ irradiates a photo sensitive material. The stopping potential was measured to be $0.5 V .$ Work function of the photo sensitive material is

1

1.68 e V

2

2.48 e V

3

0.5 e V

4

1.98 e V

Explanation:

According to Einstein's photoelectric equation,
$e V_{0} = h v - ϕ (1)$
where $V_{0}$ is the stopping potential and $ϕ$ is the work function.
Here, $V_{0} = 0.5 V, h v = 2.48 e V$
Substituting given values in equation (1), we get
$0.5 e V = 2.48 e V - ϕ$
$\Rightarrow ϕ = 1.98 e V$

JEE - 2024

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357728 Work function of nickel is $5.01 e V$ . When ultraviolet radiation of wavelength $2000 A^{\circ}$ is incident on it, electrons are emitted. What will be the maximum velecity of emitted electrons?

1

3 \times 10^{8} m / s

2

6.46 \times 10^{5} m / s

3

10.36 \times 10^{5} m / s

4

8.54 \times 10^{6} m / s

Explanation:

Energy corresponding to $2000 A^{\circ}$ .
$E = \frac{12375}{2000} e V = 6.2 e V$
Maximum kinetic energy.
$K = h v - h v_{0}$
$K = E - W_{0} \Rightarrow K_{max}$
$= (6.2 - 5.01) e V = 1.19 e V$
Now, $K_{max} = \frac{1}{2} m v^{2}_{max}$
$1.19 \times 1.6 \times 10^{- 19} = \frac{1}{2} \times 9.1 \times 10^{- 31} \times V_{max}^{2}$
$v_{max}^{2} = \frac{1.19 \times 10^{- 19} \times 1.6 \times 2}{9.1 \times 10^{- 31}}$
$= 0.418 \times 10^{12} = 418 \times 10^{10}$
$v_{max} = 6.46 \times 10^{5} m / s$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357729 The photoelectric threshold of metal is $2000$ $A^{\circ}$ . The energy of the electrons ejected from the surface by ultraviolet light of wavelength $1500$ $A^{\circ}$ is

1

2.0 e V

2

1.5 e V

3

15 e V

4

150 e V

Explanation:

$E_{k} = \frac{h c}{e} (\frac{1}{λ} - \frac{1}{λ_{0}}) = 2.0 e V$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357726 The Einstein's photoelectric equation is based upon the conservation of

1 Mass

2 Momentum

3 Angular momentum

4 Energy

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357727 In photoelectric experiment energy of $2.48 e V$ irradiates a photo sensitive material. The stopping potential was measured to be $0.5 V .$ Work function of the photo sensitive material is

1

1.68 e V

2

2.48 e V

3

0.5 e V

4

1.98 e V

Explanation:

According to Einstein's photoelectric equation,
$e V_{0} = h v - ϕ (1)$
where $V_{0}$ is the stopping potential and $ϕ$ is the work function.
Here, $V_{0} = 0.5 V, h v = 2.48 e V$
Substituting given values in equation (1), we get
$0.5 e V = 2.48 e V - ϕ$
$\Rightarrow ϕ = 1.98 e V$

JEE - 2024

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357728 Work function of nickel is $5.01 e V$ . When ultraviolet radiation of wavelength $2000 A^{\circ}$ is incident on it, electrons are emitted. What will be the maximum velecity of emitted electrons?

1

3 \times 10^{8} m / s

2

6.46 \times 10^{5} m / s

3

10.36 \times 10^{5} m / s

4

8.54 \times 10^{6} m / s

Explanation:

Energy corresponding to $2000 A^{\circ}$ .
$E = \frac{12375}{2000} e V = 6.2 e V$
Maximum kinetic energy.
$K = h v - h v_{0}$
$K = E - W_{0} \Rightarrow K_{max}$
$= (6.2 - 5.01) e V = 1.19 e V$
Now, $K_{max} = \frac{1}{2} m v^{2}_{max}$
$1.19 \times 1.6 \times 10^{- 19} = \frac{1}{2} \times 9.1 \times 10^{- 31} \times V_{max}^{2}$
$v_{max}^{2} = \frac{1.19 \times 10^{- 19} \times 1.6 \times 2}{9.1 \times 10^{- 31}}$
$= 0.418 \times 10^{12} = 418 \times 10^{10}$
$v_{max} = 6.46 \times 10^{5} m / s$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357729 The photoelectric threshold of metal is $2000$ $A^{\circ}$ . The energy of the electrons ejected from the surface by ultraviolet light of wavelength $1500$ $A^{\circ}$ is

1

2.0 e V

2

1.5 e V

3

15 e V

4

150 e V

Explanation:

$E_{k} = \frac{h c}{e} (\frac{1}{λ} - \frac{1}{λ_{0}}) = 2.0 e V$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357726 The Einstein's photoelectric equation is based upon the conservation of

1 Mass

2 Momentum

3 Angular momentum

4 Energy

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357727 In photoelectric experiment energy of $2.48 e V$ irradiates a photo sensitive material. The stopping potential was measured to be $0.5 V .$ Work function of the photo sensitive material is

1

1.68 e V

2

2.48 e V

3

0.5 e V

4

1.98 e V

Explanation:

According to Einstein's photoelectric equation,
$e V_{0} = h v - ϕ (1)$
where $V_{0}$ is the stopping potential and $ϕ$ is the work function.
Here, $V_{0} = 0.5 V, h v = 2.48 e V$
Substituting given values in equation (1), we get
$0.5 e V = 2.48 e V - ϕ$
$\Rightarrow ϕ = 1.98 e V$

JEE - 2024

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357728 Work function of nickel is $5.01 e V$ . When ultraviolet radiation of wavelength $2000 A^{\circ}$ is incident on it, electrons are emitted. What will be the maximum velecity of emitted electrons?

1

3 \times 10^{8} m / s

2

6.46 \times 10^{5} m / s

3

10.36 \times 10^{5} m / s

4

8.54 \times 10^{6} m / s

Explanation:

Energy corresponding to $2000 A^{\circ}$ .
$E = \frac{12375}{2000} e V = 6.2 e V$
Maximum kinetic energy.
$K = h v - h v_{0}$
$K = E - W_{0} \Rightarrow K_{max}$
$= (6.2 - 5.01) e V = 1.19 e V$
Now, $K_{max} = \frac{1}{2} m v^{2}_{max}$
$1.19 \times 1.6 \times 10^{- 19} = \frac{1}{2} \times 9.1 \times 10^{- 31} \times V_{max}^{2}$
$v_{max}^{2} = \frac{1.19 \times 10^{- 19} \times 1.6 \times 2}{9.1 \times 10^{- 31}}$
$= 0.418 \times 10^{12} = 418 \times 10^{10}$
$v_{max} = 6.46 \times 10^{5} m / s$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357729 The photoelectric threshold of metal is $2000$ $A^{\circ}$ . The energy of the electrons ejected from the surface by ultraviolet light of wavelength $1500$ $A^{\circ}$ is

1

2.0 e V

2

1.5 e V

3

15 e V

4

150 e V

Explanation:

$E_{k} = \frac{h c}{e} (\frac{1}{λ} - \frac{1}{λ_{0}}) = 2.0 e V$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357726 The Einstein's photoelectric equation is based upon the conservation of

1 Mass

2 Momentum

3 Angular momentum

4 Energy

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357727 In photoelectric experiment energy of $2.48 e V$ irradiates a photo sensitive material. The stopping potential was measured to be $0.5 V .$ Work function of the photo sensitive material is

1

1.68 e V

2

2.48 e V

3

0.5 e V

4

1.98 e V

Explanation:

According to Einstein's photoelectric equation,
$e V_{0} = h v - ϕ (1)$
where $V_{0}$ is the stopping potential and $ϕ$ is the work function.
Here, $V_{0} = 0.5 V, h v = 2.48 e V$
Substituting given values in equation (1), we get
$0.5 e V = 2.48 e V - ϕ$
$\Rightarrow ϕ = 1.98 e V$

JEE - 2024

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357728 Work function of nickel is $5.01 e V$ . When ultraviolet radiation of wavelength $2000 A^{\circ}$ is incident on it, electrons are emitted. What will be the maximum velecity of emitted electrons?

1

3 \times 10^{8} m / s

2

6.46 \times 10^{5} m / s

3

10.36 \times 10^{5} m / s

4

8.54 \times 10^{6} m / s

Explanation:

Energy corresponding to $2000 A^{\circ}$ .
$E = \frac{12375}{2000} e V = 6.2 e V$
Maximum kinetic energy.
$K = h v - h v_{0}$
$K = E - W_{0} \Rightarrow K_{max}$
$= (6.2 - 5.01) e V = 1.19 e V$
Now, $K_{max} = \frac{1}{2} m v^{2}_{max}$
$1.19 \times 1.6 \times 10^{- 19} = \frac{1}{2} \times 9.1 \times 10^{- 31} \times V_{max}^{2}$
$v_{max}^{2} = \frac{1.19 \times 10^{- 19} \times 1.6 \times 2}{9.1 \times 10^{- 31}}$
$= 0.418 \times 10^{12} = 418 \times 10^{10}$
$v_{max} = 6.46 \times 10^{5} m / s$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357729 The photoelectric threshold of metal is $2000$ $A^{\circ}$ . The energy of the electrons ejected from the surface by ultraviolet light of wavelength $1500$ $A^{\circ}$ is

1

2.0 e V

2

1.5 e V

3

15 e V

4

150 e V

Explanation:

$E_{k} = \frac{h c}{e} (\frac{1}{λ} - \frac{1}{λ_{0}}) = 2.0 e V$