QBManager

Dual nature of radiation and Matter

142571 The $21 cm$ radiowave emitted by hydrogen in interstellar space is due to the interaction called the hyperfine interaction in atomic hydrogen. The energy of the emitted wave is nearly

1

10^{- 17} J

2

1 J

3

7 \times 10^{- 6} J

4

10^{- 24} J

Explanation:

D Given,
$λ = 21 cm = 21 \times 10^{- 2} m$
$Energy = h v$
$= \frac{hc}{λ} = \frac{6.62 \times 10^{- 34} \times 3 \times 10^{8}}{21 \times 10^{- 2}}$
$E = 0.946 \times 10^{- 24}$
$E ◻ 1 \times 10^{- 24} J$

AIPMT - 1998

Dual nature of radiation and Matter

142572 A light source is at a distance $d$ from a photoelectric cell, then the number of photoelectrons emitted from the cell is $n$ . If the distance of light source and cell is reduced to half, then the number of photoelectrons emitted will become

1

\frac{n}{2}

2

2 n

3

4 n

4

n

Explanation:

C We know that, Intensity of light
$I \propto \frac{1}{d^{2}}$
$d =$ Distance
$\frac{I_{1}}{I_{2}} = {(\frac{d_{2}}{d_{1}})}^{2} = {(\frac{1}{2})}^{2} = \frac{1}{4}$
$I_{2} = 4 I_{1}$
$(∴ I_{1} = n)$
$I_{2} = 4 n$

AIPMT - 2001

Dual nature of radiation and Matter

142573 A photosensitive metallic surface has work function, $h v_{0}$ . If photons of energy $2 h v_{0}$ fall on this surface, the electrons come out with a maximum velocity of $4 \times 10^{6} m / s$ . When the photon energy is increased to $5 h v_{0}$ , then maximum velocity of photoelectrons will be

1

2 \times 10^{6} m / s

2

2 \times 10^{7} m / s

3

8 \times 10^{5} m / s

4

8 \times 10^{6} m / s

Explanation:

D Given,
Work function $(ϕ_{0}) = h v_{0}$
$v_{1} = 4 \times 10^{6} m / s$
$E_{1} = 2 h v_{0}$
$E_{2} = 5 h v_{0}$
Kinetic energy of photon
${KE}_{1} = E_{1} - ϕ_{0}$
$\frac{1}{2} {mv}_{1}^{2} = 2 h v_{0} - hv v_{0} = h v_{0}$
$\frac{1}{2} {mv}_{2}^{2} = 5 h v_{0} - h v_{0} = 4 hv v_{0}$
From equation (i) and (ii)
$\frac{1}{2} {mv}_{2}^{2} = 4 \times \frac{1}{2} {mv}_{1}^{2}$
$v_{2} = 2 v_{1}$
$v_{2} = 2 \times 4 \times 10^{6}$
$v_{2} = 8 \times 10^{6} m / s$

AIPMT - 2005

Dual nature of radiation and Matter

142574 The kinetic energy of an electron is $5 eV$ . Calculate the de-Broglie wavelength associated with it is.
$(Take m_{e} = 9.1 \times 10^{- 31} kg$
$and h = 6.6 \times 10^{- 34} J - s)$

1

5.47 \AA

2

10.9 \AA

3

2.7 \AA

4

3.82 \AA

Explanation:

A Given that,
We know that,
$λ = \frac{h}{\sqrt{2 mE}}$
$λ = \frac{6.6 \times 10^{- 34}}{\sqrt{2 \times 9.1 \times 10^{- 31} \times 5 \times 1.6 \times 10^{- 19}}}$
$λ = 5.469 \times 10^{- 10} m$
$λ = 5.47 \AA$

AP EMCET(Medical)-2011

Dual nature of radiation and Matter

142571 The $21 cm$ radiowave emitted by hydrogen in interstellar space is due to the interaction called the hyperfine interaction in atomic hydrogen. The energy of the emitted wave is nearly

1

10^{- 17} J

2

1 J

3

7 \times 10^{- 6} J

4

10^{- 24} J

Explanation:

D Given,
$λ = 21 cm = 21 \times 10^{- 2} m$
$Energy = h v$
$= \frac{hc}{λ} = \frac{6.62 \times 10^{- 34} \times 3 \times 10^{8}}{21 \times 10^{- 2}}$
$E = 0.946 \times 10^{- 24}$
$E ◻ 1 \times 10^{- 24} J$

AIPMT - 1998

Dual nature of radiation and Matter

142572 A light source is at a distance $d$ from a photoelectric cell, then the number of photoelectrons emitted from the cell is $n$ . If the distance of light source and cell is reduced to half, then the number of photoelectrons emitted will become

1

\frac{n}{2}

2

2 n

3

4 n

4

n

Explanation:

C We know that, Intensity of light
$I \propto \frac{1}{d^{2}}$
$d =$ Distance
$\frac{I_{1}}{I_{2}} = {(\frac{d_{2}}{d_{1}})}^{2} = {(\frac{1}{2})}^{2} = \frac{1}{4}$
$I_{2} = 4 I_{1}$
$(∴ I_{1} = n)$
$I_{2} = 4 n$

AIPMT - 2001

Dual nature of radiation and Matter

142573 A photosensitive metallic surface has work function, $h v_{0}$ . If photons of energy $2 h v_{0}$ fall on this surface, the electrons come out with a maximum velocity of $4 \times 10^{6} m / s$ . When the photon energy is increased to $5 h v_{0}$ , then maximum velocity of photoelectrons will be

1

2 \times 10^{6} m / s

2

2 \times 10^{7} m / s

3

8 \times 10^{5} m / s

4

8 \times 10^{6} m / s

Explanation:

D Given,
Work function $(ϕ_{0}) = h v_{0}$
$v_{1} = 4 \times 10^{6} m / s$
$E_{1} = 2 h v_{0}$
$E_{2} = 5 h v_{0}$
Kinetic energy of photon
${KE}_{1} = E_{1} - ϕ_{0}$
$\frac{1}{2} {mv}_{1}^{2} = 2 h v_{0} - hv v_{0} = h v_{0}$
$\frac{1}{2} {mv}_{2}^{2} = 5 h v_{0} - h v_{0} = 4 hv v_{0}$
From equation (i) and (ii)
$\frac{1}{2} {mv}_{2}^{2} = 4 \times \frac{1}{2} {mv}_{1}^{2}$
$v_{2} = 2 v_{1}$
$v_{2} = 2 \times 4 \times 10^{6}$
$v_{2} = 8 \times 10^{6} m / s$

AIPMT - 2005

Dual nature of radiation and Matter

142574 The kinetic energy of an electron is $5 eV$ . Calculate the de-Broglie wavelength associated with it is.
$(Take m_{e} = 9.1 \times 10^{- 31} kg$
$and h = 6.6 \times 10^{- 34} J - s)$

1

5.47 \AA

2

10.9 \AA

3

2.7 \AA

4

3.82 \AA

Explanation:

A Given that,
We know that,
$λ = \frac{h}{\sqrt{2 mE}}$
$λ = \frac{6.6 \times 10^{- 34}}{\sqrt{2 \times 9.1 \times 10^{- 31} \times 5 \times 1.6 \times 10^{- 19}}}$
$λ = 5.469 \times 10^{- 10} m$
$λ = 5.47 \AA$

AP EMCET(Medical)-2011

Dual nature of radiation and Matter

142571 The $21 cm$ radiowave emitted by hydrogen in interstellar space is due to the interaction called the hyperfine interaction in atomic hydrogen. The energy of the emitted wave is nearly

1

10^{- 17} J

2

1 J

3

7 \times 10^{- 6} J

4

10^{- 24} J

Explanation:

D Given,
$λ = 21 cm = 21 \times 10^{- 2} m$
$Energy = h v$
$= \frac{hc}{λ} = \frac{6.62 \times 10^{- 34} \times 3 \times 10^{8}}{21 \times 10^{- 2}}$
$E = 0.946 \times 10^{- 24}$
$E ◻ 1 \times 10^{- 24} J$

AIPMT - 1998

Dual nature of radiation and Matter

142572 A light source is at a distance $d$ from a photoelectric cell, then the number of photoelectrons emitted from the cell is $n$ . If the distance of light source and cell is reduced to half, then the number of photoelectrons emitted will become

1

\frac{n}{2}

2

2 n

3

4 n

4

n

Explanation:

C We know that, Intensity of light
$I \propto \frac{1}{d^{2}}$
$d =$ Distance
$\frac{I_{1}}{I_{2}} = {(\frac{d_{2}}{d_{1}})}^{2} = {(\frac{1}{2})}^{2} = \frac{1}{4}$
$I_{2} = 4 I_{1}$
$(∴ I_{1} = n)$
$I_{2} = 4 n$

AIPMT - 2001

Dual nature of radiation and Matter

142573 A photosensitive metallic surface has work function, $h v_{0}$ . If photons of energy $2 h v_{0}$ fall on this surface, the electrons come out with a maximum velocity of $4 \times 10^{6} m / s$ . When the photon energy is increased to $5 h v_{0}$ , then maximum velocity of photoelectrons will be

1

2 \times 10^{6} m / s

2

2 \times 10^{7} m / s

3

8 \times 10^{5} m / s

4

8 \times 10^{6} m / s

Explanation:

D Given,
Work function $(ϕ_{0}) = h v_{0}$
$v_{1} = 4 \times 10^{6} m / s$
$E_{1} = 2 h v_{0}$
$E_{2} = 5 h v_{0}$
Kinetic energy of photon
${KE}_{1} = E_{1} - ϕ_{0}$
$\frac{1}{2} {mv}_{1}^{2} = 2 h v_{0} - hv v_{0} = h v_{0}$
$\frac{1}{2} {mv}_{2}^{2} = 5 h v_{0} - h v_{0} = 4 hv v_{0}$
From equation (i) and (ii)
$\frac{1}{2} {mv}_{2}^{2} = 4 \times \frac{1}{2} {mv}_{1}^{2}$
$v_{2} = 2 v_{1}$
$v_{2} = 2 \times 4 \times 10^{6}$
$v_{2} = 8 \times 10^{6} m / s$

AIPMT - 2005

Dual nature of radiation and Matter

142574 The kinetic energy of an electron is $5 eV$ . Calculate the de-Broglie wavelength associated with it is.
$(Take m_{e} = 9.1 \times 10^{- 31} kg$
$and h = 6.6 \times 10^{- 34} J - s)$

1

5.47 \AA

2

10.9 \AA

3

2.7 \AA

4

3.82 \AA

Explanation:

A Given that,
We know that,
$λ = \frac{h}{\sqrt{2 mE}}$
$λ = \frac{6.6 \times 10^{- 34}}{\sqrt{2 \times 9.1 \times 10^{- 31} \times 5 \times 1.6 \times 10^{- 19}}}$
$λ = 5.469 \times 10^{- 10} m$
$λ = 5.47 \AA$

AP EMCET(Medical)-2011

Dual nature of radiation and Matter

142571 The $21 cm$ radiowave emitted by hydrogen in interstellar space is due to the interaction called the hyperfine interaction in atomic hydrogen. The energy of the emitted wave is nearly

1

10^{- 17} J

2

1 J

3

7 \times 10^{- 6} J

4

10^{- 24} J

Explanation:

D Given,
$λ = 21 cm = 21 \times 10^{- 2} m$
$Energy = h v$
$= \frac{hc}{λ} = \frac{6.62 \times 10^{- 34} \times 3 \times 10^{8}}{21 \times 10^{- 2}}$
$E = 0.946 \times 10^{- 24}$
$E ◻ 1 \times 10^{- 24} J$

AIPMT - 1998

Dual nature of radiation and Matter

142572 A light source is at a distance $d$ from a photoelectric cell, then the number of photoelectrons emitted from the cell is $n$ . If the distance of light source and cell is reduced to half, then the number of photoelectrons emitted will become

1

\frac{n}{2}

2

2 n

3

4 n

4

n

Explanation:

C We know that, Intensity of light
$I \propto \frac{1}{d^{2}}$
$d =$ Distance
$\frac{I_{1}}{I_{2}} = {(\frac{d_{2}}{d_{1}})}^{2} = {(\frac{1}{2})}^{2} = \frac{1}{4}$
$I_{2} = 4 I_{1}$
$(∴ I_{1} = n)$
$I_{2} = 4 n$

AIPMT - 2001

Dual nature of radiation and Matter

142573 A photosensitive metallic surface has work function, $h v_{0}$ . If photons of energy $2 h v_{0}$ fall on this surface, the electrons come out with a maximum velocity of $4 \times 10^{6} m / s$ . When the photon energy is increased to $5 h v_{0}$ , then maximum velocity of photoelectrons will be

1

2 \times 10^{6} m / s

2

2 \times 10^{7} m / s

3

8 \times 10^{5} m / s

4

8 \times 10^{6} m / s

Explanation:

D Given,
Work function $(ϕ_{0}) = h v_{0}$
$v_{1} = 4 \times 10^{6} m / s$
$E_{1} = 2 h v_{0}$
$E_{2} = 5 h v_{0}$
Kinetic energy of photon
${KE}_{1} = E_{1} - ϕ_{0}$
$\frac{1}{2} {mv}_{1}^{2} = 2 h v_{0} - hv v_{0} = h v_{0}$
$\frac{1}{2} {mv}_{2}^{2} = 5 h v_{0} - h v_{0} = 4 hv v_{0}$
From equation (i) and (ii)
$\frac{1}{2} {mv}_{2}^{2} = 4 \times \frac{1}{2} {mv}_{1}^{2}$
$v_{2} = 2 v_{1}$
$v_{2} = 2 \times 4 \times 10^{6}$
$v_{2} = 8 \times 10^{6} m / s$

AIPMT - 2005

Dual nature of radiation and Matter

142574 The kinetic energy of an electron is $5 eV$ . Calculate the de-Broglie wavelength associated with it is.
$(Take m_{e} = 9.1 \times 10^{- 31} kg$
$and h = 6.6 \times 10^{- 34} J - s)$

1

5.47 \AA

2

10.9 \AA

3

2.7 \AA

4

3.82 \AA

Explanation:

A Given that,
We know that,
$λ = \frac{h}{\sqrt{2 mE}}$
$λ = \frac{6.6 \times 10^{- 34}}{\sqrt{2 \times 9.1 \times 10^{- 31} \times 5 \times 1.6 \times 10^{- 19}}}$
$λ = 5.469 \times 10^{- 10} m$
$λ = 5.47 \AA$

AP EMCET(Medical)-2011

Explanation:

Explanation:

Explanation:

Explanation:

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Question Bank Series