QBManager

Dual nature of radiation and Matter

142576 The de-Broglie wavelength of a particle moving with a velocity $2.25 \times 10^{8} m / s$ is equal to the wavelength of a photon. The ratio of kinetic energy of the photon. The ratio of kinetic energy of the particle to the energy of the photon is :
(Velocity of light is $3 \times 10^{8} m / s$ )

1

\frac{1}{8}

2

\frac{3}{8}

3

\frac{5}{8}

4

\frac{7}{8}

Explanation:

B Given,
$v = 2.25 \times 10^{8} m / s$
$c = 3 \times 10^{8} m / s$
We know that,
$K_{photon} = \frac{hc}{λ}$
$K_{particle} = \frac{1}{2} \times m \cdot v^{2}$
Here,
$λ = \frac{h}{mv}$
$m = \frac{h}{λ \cdot v}$
$K_{particle} = \frac{1}{2} \times (\frac{h}{λ v}) v^{2}$
$K_{particle} = \frac{hv}{2 λ}$
Now, the ratio are -
$\frac{K_{particle}}{K_{photon}} = \frac{\frac{hv}{2 λ}}{\frac{hc}{λ}} = \frac{v}{2 c}$
$\frac{K_{particle}}{K_{photon}} = \frac{2.25 \times 10^{8}}{2 \times 3 \times 10^{8}}$
$\frac{K_{particle}}{K_{photon}} = \frac{3}{8}$

AP EAMCET(Medical)-2003

Dual nature of radiation and Matter

142577 The value of de-Broglie wavelength of an electron moving with a speed of $6.6 \times 10^{5} m / s$ is approximately
[Planck's constant $h = 6.6 \times 10^{- 34} J - s$ mass of electron $= 9 \times 10^{- 31} kg$ ]

1

11 \AA

2

111 \AA

3

211 \AA

4

311 \AA

Explanation:

A Given,
$h = 6.6 \times 10^{- 34} J s$
$m = 9.1 \times 10^{- 31} kg$
$v = 6.6 \times 10^{5} m / s$
We know that,
$λ = \frac{h}{mv} = \frac{6.6 \times 10^{- 34}}{9.1 \times 10^{- 31} \times 6.6 \times 10^{5}}$
$λ = 11 \AA$

AP EAMCET(Medical)-2002

Dual nature of radiation and Matter

142578 Electrons accelerated by a potential of $V$ volt strike a target material to produce continuous X-rays. Ratio between the de-Broglie wavelength of the electrons striking the target and the shortest wavelength of the continuous $X$ -rays emitted is

1

\frac{h}{\sqrt{2 Vem}}

2

\frac{1}{c} \sqrt{\frac{2 m}{Ve}}

3

\frac{1}{c} \sqrt{\frac{Ve}{2 m}}

4

\frac{hc}{\sqrt{\frac{Ve}{2 m}}}

Explanation:

C Shortest wavelength of the continuous X Ray emitted
$E_{1} = \frac{hc}{λ_{1}}$
$eV = \frac{hc}{λ_{1}}$
$λ_{1} = (\frac{hc}{eV})$
de - Broglie wavelength of electron striking the target
$E_{2} = \frac{h}{mv}$
$\frac{1}{2} {mv}^{2} = eV$
$v^{2} = (\frac{2 eV}{m})$
$v = \sqrt{\frac{2 eV}{m}}$
$∴ λ_{2} = \frac{h}{m \sqrt{\frac{2 eV}{m}}}$
$λ_{2} = \frac{h}{\sqrt{2 mVe}}$
Dividing equation (ii) by (i), we have-
$\frac{λ_{2}}{λ_{1}} = \frac{h / \sqrt{2 meV}}{hc / eV}$
$\frac{λ_{2}}{λ_{1}} = \frac{1}{c} \sqrt{\frac{eV}{2 m}}$

AP EAMCET(Medical)-2009

Dual nature of radiation and Matter

142579 The de-Broglie wavelength of an electron having $80 eV$ energy is nearly
$(1 eV = 1.6 \times 10^{- 19} J)$
Mass of the electron $= 9 \times 10^{- 31} kg$
Planck's constant $= 6.6 \times 10^{- 34} J - s$

1

140 \AA

2

0.14 \AA

3

14 \AA

4

1.4 \AA

Explanation:

D Given,
$1 eV = 1.6 \times 10^{- 19} J$
$m = 9 \times 10^{- 31} kg$
$h = 6.6 \times 10^{- 34} J - s$
We know that,
$λ = \frac{h}{\sqrt{2 mE}}$
$λ = \frac{6.6 \times 10^{- 34}}{\sqrt{2 \times 9 \times 10^{- 31} \times 80 \times 1.6 \times 10^{- 19}}}$
$λ = 1.375 \times 10^{- 10} m$
$λ = 1.4 \AA$

EAMCET-2001

Dual nature of radiation and Matter

142580 In Compton scattering process, the incident $X$ radiation is scattered at an angle $60^{\circ}$ . The wavelength of the scattered radiation is $0.22 \AA$ . The wavelength of the incident $X$ -radiation in $\AA$ units is

1 0.508

2 0.408

3 0.232

4 0.208

Explanation:

D Given
$θ = 60^{\circ}$
Wavelength of striking radiation $(λ^{'}) = 0.22 \AA$ Wavelength of the incident $X$ - radiation $(λ) =$ ?
$λ^{'} = λ + \frac{h}{m_{e} c} [1 - \cos θ]$
$λ = λ^{'} - \frac{h}{m_{e} c} [1 - \frac{1}{2}]$
$λ = (0.22 \times 10^{- 10}) - \frac{6.6 \times 10^{- 34}}{9.1 \times 10^{- 31} \times 3 \times 10^{8}} \times \frac{1}{2}$
$λ = (0.220 - 0.012) \times 10^{- 10} m$
$λ = 0.208 \AA$

EAMCET-2002

Dual nature of radiation and Matter

142576 The de-Broglie wavelength of a particle moving with a velocity $2.25 \times 10^{8} m / s$ is equal to the wavelength of a photon. The ratio of kinetic energy of the photon. The ratio of kinetic energy of the particle to the energy of the photon is :
(Velocity of light is $3 \times 10^{8} m / s$ )

1

\frac{1}{8}

2

\frac{3}{8}

3

\frac{5}{8}

4

\frac{7}{8}

Explanation:

B Given,
$v = 2.25 \times 10^{8} m / s$
$c = 3 \times 10^{8} m / s$
We know that,
$K_{photon} = \frac{hc}{λ}$
$K_{particle} = \frac{1}{2} \times m \cdot v^{2}$
Here,
$λ = \frac{h}{mv}$
$m = \frac{h}{λ \cdot v}$
$K_{particle} = \frac{1}{2} \times (\frac{h}{λ v}) v^{2}$
$K_{particle} = \frac{hv}{2 λ}$
Now, the ratio are -
$\frac{K_{particle}}{K_{photon}} = \frac{\frac{hv}{2 λ}}{\frac{hc}{λ}} = \frac{v}{2 c}$
$\frac{K_{particle}}{K_{photon}} = \frac{2.25 \times 10^{8}}{2 \times 3 \times 10^{8}}$
$\frac{K_{particle}}{K_{photon}} = \frac{3}{8}$

AP EAMCET(Medical)-2003

Dual nature of radiation and Matter

142577 The value of de-Broglie wavelength of an electron moving with a speed of $6.6 \times 10^{5} m / s$ is approximately
[Planck's constant $h = 6.6 \times 10^{- 34} J - s$ mass of electron $= 9 \times 10^{- 31} kg$ ]

1

11 \AA

2

111 \AA

3

211 \AA

4

311 \AA

Explanation:

A Given,
$h = 6.6 \times 10^{- 34} J s$
$m = 9.1 \times 10^{- 31} kg$
$v = 6.6 \times 10^{5} m / s$
We know that,
$λ = \frac{h}{mv} = \frac{6.6 \times 10^{- 34}}{9.1 \times 10^{- 31} \times 6.6 \times 10^{5}}$
$λ = 11 \AA$

AP EAMCET(Medical)-2002

Dual nature of radiation and Matter

142578 Electrons accelerated by a potential of $V$ volt strike a target material to produce continuous X-rays. Ratio between the de-Broglie wavelength of the electrons striking the target and the shortest wavelength of the continuous $X$ -rays emitted is

1

\frac{h}{\sqrt{2 Vem}}

2

\frac{1}{c} \sqrt{\frac{2 m}{Ve}}

3

\frac{1}{c} \sqrt{\frac{Ve}{2 m}}

4

\frac{hc}{\sqrt{\frac{Ve}{2 m}}}

Explanation:

C Shortest wavelength of the continuous X Ray emitted
$E_{1} = \frac{hc}{λ_{1}}$
$eV = \frac{hc}{λ_{1}}$
$λ_{1} = (\frac{hc}{eV})$
de - Broglie wavelength of electron striking the target
$E_{2} = \frac{h}{mv}$
$\frac{1}{2} {mv}^{2} = eV$
$v^{2} = (\frac{2 eV}{m})$
$v = \sqrt{\frac{2 eV}{m}}$
$∴ λ_{2} = \frac{h}{m \sqrt{\frac{2 eV}{m}}}$
$λ_{2} = \frac{h}{\sqrt{2 mVe}}$
Dividing equation (ii) by (i), we have-
$\frac{λ_{2}}{λ_{1}} = \frac{h / \sqrt{2 meV}}{hc / eV}$
$\frac{λ_{2}}{λ_{1}} = \frac{1}{c} \sqrt{\frac{eV}{2 m}}$

AP EAMCET(Medical)-2009

Dual nature of radiation and Matter

142579 The de-Broglie wavelength of an electron having $80 eV$ energy is nearly
$(1 eV = 1.6 \times 10^{- 19} J)$
Mass of the electron $= 9 \times 10^{- 31} kg$
Planck's constant $= 6.6 \times 10^{- 34} J - s$

1

140 \AA

2

0.14 \AA

3

14 \AA

4

1.4 \AA

Explanation:

D Given,
$1 eV = 1.6 \times 10^{- 19} J$
$m = 9 \times 10^{- 31} kg$
$h = 6.6 \times 10^{- 34} J - s$
We know that,
$λ = \frac{h}{\sqrt{2 mE}}$
$λ = \frac{6.6 \times 10^{- 34}}{\sqrt{2 \times 9 \times 10^{- 31} \times 80 \times 1.6 \times 10^{- 19}}}$
$λ = 1.375 \times 10^{- 10} m$
$λ = 1.4 \AA$

EAMCET-2001

Dual nature of radiation and Matter

142580 In Compton scattering process, the incident $X$ radiation is scattered at an angle $60^{\circ}$ . The wavelength of the scattered radiation is $0.22 \AA$ . The wavelength of the incident $X$ -radiation in $\AA$ units is

1 0.508

2 0.408

3 0.232

4 0.208

Explanation:

D Given
$θ = 60^{\circ}$
Wavelength of striking radiation $(λ^{'}) = 0.22 \AA$ Wavelength of the incident $X$ - radiation $(λ) =$ ?
$λ^{'} = λ + \frac{h}{m_{e} c} [1 - \cos θ]$
$λ = λ^{'} - \frac{h}{m_{e} c} [1 - \frac{1}{2}]$
$λ = (0.22 \times 10^{- 10}) - \frac{6.6 \times 10^{- 34}}{9.1 \times 10^{- 31} \times 3 \times 10^{8}} \times \frac{1}{2}$
$λ = (0.220 - 0.012) \times 10^{- 10} m$
$λ = 0.208 \AA$

EAMCET-2002

Dual nature of radiation and Matter

142576 The de-Broglie wavelength of a particle moving with a velocity $2.25 \times 10^{8} m / s$ is equal to the wavelength of a photon. The ratio of kinetic energy of the photon. The ratio of kinetic energy of the particle to the energy of the photon is :
(Velocity of light is $3 \times 10^{8} m / s$ )

1

\frac{1}{8}

2

\frac{3}{8}

3

\frac{5}{8}

4

\frac{7}{8}

Explanation:

B Given,
$v = 2.25 \times 10^{8} m / s$
$c = 3 \times 10^{8} m / s$
We know that,
$K_{photon} = \frac{hc}{λ}$
$K_{particle} = \frac{1}{2} \times m \cdot v^{2}$
Here,
$λ = \frac{h}{mv}$
$m = \frac{h}{λ \cdot v}$
$K_{particle} = \frac{1}{2} \times (\frac{h}{λ v}) v^{2}$
$K_{particle} = \frac{hv}{2 λ}$
Now, the ratio are -
$\frac{K_{particle}}{K_{photon}} = \frac{\frac{hv}{2 λ}}{\frac{hc}{λ}} = \frac{v}{2 c}$
$\frac{K_{particle}}{K_{photon}} = \frac{2.25 \times 10^{8}}{2 \times 3 \times 10^{8}}$
$\frac{K_{particle}}{K_{photon}} = \frac{3}{8}$

AP EAMCET(Medical)-2003

Dual nature of radiation and Matter

142577 The value of de-Broglie wavelength of an electron moving with a speed of $6.6 \times 10^{5} m / s$ is approximately
[Planck's constant $h = 6.6 \times 10^{- 34} J - s$ mass of electron $= 9 \times 10^{- 31} kg$ ]

1

11 \AA

2

111 \AA

3

211 \AA

4

311 \AA

Explanation:

A Given,
$h = 6.6 \times 10^{- 34} J s$
$m = 9.1 \times 10^{- 31} kg$
$v = 6.6 \times 10^{5} m / s$
We know that,
$λ = \frac{h}{mv} = \frac{6.6 \times 10^{- 34}}{9.1 \times 10^{- 31} \times 6.6 \times 10^{5}}$
$λ = 11 \AA$

AP EAMCET(Medical)-2002

Dual nature of radiation and Matter

142578 Electrons accelerated by a potential of $V$ volt strike a target material to produce continuous X-rays. Ratio between the de-Broglie wavelength of the electrons striking the target and the shortest wavelength of the continuous $X$ -rays emitted is

1

\frac{h}{\sqrt{2 Vem}}

2

\frac{1}{c} \sqrt{\frac{2 m}{Ve}}

3

\frac{1}{c} \sqrt{\frac{Ve}{2 m}}

4

\frac{hc}{\sqrt{\frac{Ve}{2 m}}}

Explanation:

C Shortest wavelength of the continuous X Ray emitted
$E_{1} = \frac{hc}{λ_{1}}$
$eV = \frac{hc}{λ_{1}}$
$λ_{1} = (\frac{hc}{eV})$
de - Broglie wavelength of electron striking the target
$E_{2} = \frac{h}{mv}$
$\frac{1}{2} {mv}^{2} = eV$
$v^{2} = (\frac{2 eV}{m})$
$v = \sqrt{\frac{2 eV}{m}}$
$∴ λ_{2} = \frac{h}{m \sqrt{\frac{2 eV}{m}}}$
$λ_{2} = \frac{h}{\sqrt{2 mVe}}$
Dividing equation (ii) by (i), we have-
$\frac{λ_{2}}{λ_{1}} = \frac{h / \sqrt{2 meV}}{hc / eV}$
$\frac{λ_{2}}{λ_{1}} = \frac{1}{c} \sqrt{\frac{eV}{2 m}}$

AP EAMCET(Medical)-2009

Dual nature of radiation and Matter

142579 The de-Broglie wavelength of an electron having $80 eV$ energy is nearly
$(1 eV = 1.6 \times 10^{- 19} J)$
Mass of the electron $= 9 \times 10^{- 31} kg$
Planck's constant $= 6.6 \times 10^{- 34} J - s$

1

140 \AA

2

0.14 \AA

3

14 \AA

4

1.4 \AA

Explanation:

D Given,
$1 eV = 1.6 \times 10^{- 19} J$
$m = 9 \times 10^{- 31} kg$
$h = 6.6 \times 10^{- 34} J - s$
We know that,
$λ = \frac{h}{\sqrt{2 mE}}$
$λ = \frac{6.6 \times 10^{- 34}}{\sqrt{2 \times 9 \times 10^{- 31} \times 80 \times 1.6 \times 10^{- 19}}}$
$λ = 1.375 \times 10^{- 10} m$
$λ = 1.4 \AA$

EAMCET-2001

Dual nature of radiation and Matter

142580 In Compton scattering process, the incident $X$ radiation is scattered at an angle $60^{\circ}$ . The wavelength of the scattered radiation is $0.22 \AA$ . The wavelength of the incident $X$ -radiation in $\AA$ units is

1 0.508

2 0.408

3 0.232

4 0.208

Explanation:

D Given
$θ = 60^{\circ}$
Wavelength of striking radiation $(λ^{'}) = 0.22 \AA$ Wavelength of the incident $X$ - radiation $(λ) =$ ?
$λ^{'} = λ + \frac{h}{m_{e} c} [1 - \cos θ]$
$λ = λ^{'} - \frac{h}{m_{e} c} [1 - \frac{1}{2}]$
$λ = (0.22 \times 10^{- 10}) - \frac{6.6 \times 10^{- 34}}{9.1 \times 10^{- 31} \times 3 \times 10^{8}} \times \frac{1}{2}$
$λ = (0.220 - 0.012) \times 10^{- 10} m$
$λ = 0.208 \AA$

EAMCET-2002

Dual nature of radiation and Matter

142576 The de-Broglie wavelength of a particle moving with a velocity $2.25 \times 10^{8} m / s$ is equal to the wavelength of a photon. The ratio of kinetic energy of the photon. The ratio of kinetic energy of the particle to the energy of the photon is :
(Velocity of light is $3 \times 10^{8} m / s$ )

1

\frac{1}{8}

2

\frac{3}{8}

3

\frac{5}{8}

4

\frac{7}{8}

Explanation:

B Given,
$v = 2.25 \times 10^{8} m / s$
$c = 3 \times 10^{8} m / s$
We know that,
$K_{photon} = \frac{hc}{λ}$
$K_{particle} = \frac{1}{2} \times m \cdot v^{2}$
Here,
$λ = \frac{h}{mv}$
$m = \frac{h}{λ \cdot v}$
$K_{particle} = \frac{1}{2} \times (\frac{h}{λ v}) v^{2}$
$K_{particle} = \frac{hv}{2 λ}$
Now, the ratio are -
$\frac{K_{particle}}{K_{photon}} = \frac{\frac{hv}{2 λ}}{\frac{hc}{λ}} = \frac{v}{2 c}$
$\frac{K_{particle}}{K_{photon}} = \frac{2.25 \times 10^{8}}{2 \times 3 \times 10^{8}}$
$\frac{K_{particle}}{K_{photon}} = \frac{3}{8}$

AP EAMCET(Medical)-2003

Dual nature of radiation and Matter

142577 The value of de-Broglie wavelength of an electron moving with a speed of $6.6 \times 10^{5} m / s$ is approximately
[Planck's constant $h = 6.6 \times 10^{- 34} J - s$ mass of electron $= 9 \times 10^{- 31} kg$ ]

1

11 \AA

2

111 \AA

3

211 \AA

4

311 \AA

Explanation:

A Given,
$h = 6.6 \times 10^{- 34} J s$
$m = 9.1 \times 10^{- 31} kg$
$v = 6.6 \times 10^{5} m / s$
We know that,
$λ = \frac{h}{mv} = \frac{6.6 \times 10^{- 34}}{9.1 \times 10^{- 31} \times 6.6 \times 10^{5}}$
$λ = 11 \AA$

AP EAMCET(Medical)-2002

Dual nature of radiation and Matter

142578 Electrons accelerated by a potential of $V$ volt strike a target material to produce continuous X-rays. Ratio between the de-Broglie wavelength of the electrons striking the target and the shortest wavelength of the continuous $X$ -rays emitted is

1

\frac{h}{\sqrt{2 Vem}}

2

\frac{1}{c} \sqrt{\frac{2 m}{Ve}}

3

\frac{1}{c} \sqrt{\frac{Ve}{2 m}}

4

\frac{hc}{\sqrt{\frac{Ve}{2 m}}}

Explanation:

C Shortest wavelength of the continuous X Ray emitted
$E_{1} = \frac{hc}{λ_{1}}$
$eV = \frac{hc}{λ_{1}}$
$λ_{1} = (\frac{hc}{eV})$
de - Broglie wavelength of electron striking the target
$E_{2} = \frac{h}{mv}$
$\frac{1}{2} {mv}^{2} = eV$
$v^{2} = (\frac{2 eV}{m})$
$v = \sqrt{\frac{2 eV}{m}}$
$∴ λ_{2} = \frac{h}{m \sqrt{\frac{2 eV}{m}}}$
$λ_{2} = \frac{h}{\sqrt{2 mVe}}$
Dividing equation (ii) by (i), we have-
$\frac{λ_{2}}{λ_{1}} = \frac{h / \sqrt{2 meV}}{hc / eV}$
$\frac{λ_{2}}{λ_{1}} = \frac{1}{c} \sqrt{\frac{eV}{2 m}}$

AP EAMCET(Medical)-2009

Dual nature of radiation and Matter

142579 The de-Broglie wavelength of an electron having $80 eV$ energy is nearly
$(1 eV = 1.6 \times 10^{- 19} J)$
Mass of the electron $= 9 \times 10^{- 31} kg$
Planck's constant $= 6.6 \times 10^{- 34} J - s$

1

140 \AA

2

0.14 \AA

3

14 \AA

4

1.4 \AA

Explanation:

D Given,
$1 eV = 1.6 \times 10^{- 19} J$
$m = 9 \times 10^{- 31} kg$
$h = 6.6 \times 10^{- 34} J - s$
We know that,
$λ = \frac{h}{\sqrt{2 mE}}$
$λ = \frac{6.6 \times 10^{- 34}}{\sqrt{2 \times 9 \times 10^{- 31} \times 80 \times 1.6 \times 10^{- 19}}}$
$λ = 1.375 \times 10^{- 10} m$
$λ = 1.4 \AA$

EAMCET-2001

Dual nature of radiation and Matter

142580 In Compton scattering process, the incident $X$ radiation is scattered at an angle $60^{\circ}$ . The wavelength of the scattered radiation is $0.22 \AA$ . The wavelength of the incident $X$ -radiation in $\AA$ units is

1 0.508

2 0.408

3 0.232

4 0.208

Explanation:

D Given
$θ = 60^{\circ}$
Wavelength of striking radiation $(λ^{'}) = 0.22 \AA$ Wavelength of the incident $X$ - radiation $(λ) =$ ?
$λ^{'} = λ + \frac{h}{m_{e} c} [1 - \cos θ]$
$λ = λ^{'} - \frac{h}{m_{e} c} [1 - \frac{1}{2}]$
$λ = (0.22 \times 10^{- 10}) - \frac{6.6 \times 10^{- 34}}{9.1 \times 10^{- 31} \times 3 \times 10^{8}} \times \frac{1}{2}$
$λ = (0.220 - 0.012) \times 10^{- 10} m$
$λ = 0.208 \AA$

EAMCET-2002

Dual nature of radiation and Matter

142576 The de-Broglie wavelength of a particle moving with a velocity $2.25 \times 10^{8} m / s$ is equal to the wavelength of a photon. The ratio of kinetic energy of the photon. The ratio of kinetic energy of the particle to the energy of the photon is :
(Velocity of light is $3 \times 10^{8} m / s$ )

1

\frac{1}{8}

2

\frac{3}{8}

3

\frac{5}{8}

4

\frac{7}{8}

Explanation:

B Given,
$v = 2.25 \times 10^{8} m / s$
$c = 3 \times 10^{8} m / s$
We know that,
$K_{photon} = \frac{hc}{λ}$
$K_{particle} = \frac{1}{2} \times m \cdot v^{2}$
Here,
$λ = \frac{h}{mv}$
$m = \frac{h}{λ \cdot v}$
$K_{particle} = \frac{1}{2} \times (\frac{h}{λ v}) v^{2}$
$K_{particle} = \frac{hv}{2 λ}$
Now, the ratio are -
$\frac{K_{particle}}{K_{photon}} = \frac{\frac{hv}{2 λ}}{\frac{hc}{λ}} = \frac{v}{2 c}$
$\frac{K_{particle}}{K_{photon}} = \frac{2.25 \times 10^{8}}{2 \times 3 \times 10^{8}}$
$\frac{K_{particle}}{K_{photon}} = \frac{3}{8}$

AP EAMCET(Medical)-2003

Dual nature of radiation and Matter

142577 The value of de-Broglie wavelength of an electron moving with a speed of $6.6 \times 10^{5} m / s$ is approximately
[Planck's constant $h = 6.6 \times 10^{- 34} J - s$ mass of electron $= 9 \times 10^{- 31} kg$ ]

1

11 \AA

2

111 \AA

3

211 \AA

4

311 \AA

Explanation:

A Given,
$h = 6.6 \times 10^{- 34} J s$
$m = 9.1 \times 10^{- 31} kg$
$v = 6.6 \times 10^{5} m / s$
We know that,
$λ = \frac{h}{mv} = \frac{6.6 \times 10^{- 34}}{9.1 \times 10^{- 31} \times 6.6 \times 10^{5}}$
$λ = 11 \AA$

AP EAMCET(Medical)-2002

Dual nature of radiation and Matter

142578 Electrons accelerated by a potential of $V$ volt strike a target material to produce continuous X-rays. Ratio between the de-Broglie wavelength of the electrons striking the target and the shortest wavelength of the continuous $X$ -rays emitted is

1

\frac{h}{\sqrt{2 Vem}}

2

\frac{1}{c} \sqrt{\frac{2 m}{Ve}}

3

\frac{1}{c} \sqrt{\frac{Ve}{2 m}}

4

\frac{hc}{\sqrt{\frac{Ve}{2 m}}}

Explanation:

C Shortest wavelength of the continuous X Ray emitted
$E_{1} = \frac{hc}{λ_{1}}$
$eV = \frac{hc}{λ_{1}}$
$λ_{1} = (\frac{hc}{eV})$
de - Broglie wavelength of electron striking the target
$E_{2} = \frac{h}{mv}$
$\frac{1}{2} {mv}^{2} = eV$
$v^{2} = (\frac{2 eV}{m})$
$v = \sqrt{\frac{2 eV}{m}}$
$∴ λ_{2} = \frac{h}{m \sqrt{\frac{2 eV}{m}}}$
$λ_{2} = \frac{h}{\sqrt{2 mVe}}$
Dividing equation (ii) by (i), we have-
$\frac{λ_{2}}{λ_{1}} = \frac{h / \sqrt{2 meV}}{hc / eV}$
$\frac{λ_{2}}{λ_{1}} = \frac{1}{c} \sqrt{\frac{eV}{2 m}}$

AP EAMCET(Medical)-2009

Dual nature of radiation and Matter

142579 The de-Broglie wavelength of an electron having $80 eV$ energy is nearly
$(1 eV = 1.6 \times 10^{- 19} J)$
Mass of the electron $= 9 \times 10^{- 31} kg$
Planck's constant $= 6.6 \times 10^{- 34} J - s$

1

140 \AA

2

0.14 \AA

3

14 \AA

4

1.4 \AA

Explanation:

D Given,
$1 eV = 1.6 \times 10^{- 19} J$
$m = 9 \times 10^{- 31} kg$
$h = 6.6 \times 10^{- 34} J - s$
We know that,
$λ = \frac{h}{\sqrt{2 mE}}$
$λ = \frac{6.6 \times 10^{- 34}}{\sqrt{2 \times 9 \times 10^{- 31} \times 80 \times 1.6 \times 10^{- 19}}}$
$λ = 1.375 \times 10^{- 10} m$
$λ = 1.4 \AA$

EAMCET-2001

Dual nature of radiation and Matter

142580 In Compton scattering process, the incident $X$ radiation is scattered at an angle $60^{\circ}$ . The wavelength of the scattered radiation is $0.22 \AA$ . The wavelength of the incident $X$ -radiation in $\AA$ units is

1 0.508

2 0.408

3 0.232

4 0.208

Explanation:

D Given
$θ = 60^{\circ}$
Wavelength of striking radiation $(λ^{'}) = 0.22 \AA$ Wavelength of the incident $X$ - radiation $(λ) =$ ?
$λ^{'} = λ + \frac{h}{m_{e} c} [1 - \cos θ]$
$λ = λ^{'} - \frac{h}{m_{e} c} [1 - \frac{1}{2}]$
$λ = (0.22 \times 10^{- 10}) - \frac{6.6 \times 10^{- 34}}{9.1 \times 10^{- 31} \times 3 \times 10^{8}} \times \frac{1}{2}$
$λ = (0.220 - 0.012) \times 10^{- 10} m$
$λ = 0.208 \AA$

EAMCET-2002

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Question Bank Series

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