QBManager

Dual nature of radiation and Matter

142560 Electrons of mass $m$ with de-Broglie wavelength
$λ$ fall on the target in an X-ray tube. The cutoff wavelength $(λ_{0})$ of the emitted $X$ -rays is

1

λ_{0} = \frac{2 mc λ^{2}}{h}

2

λ_{0} = \frac{2 h}{mc}

3

λ_{0} = \frac{2 m^{2} c^{2} λ^{3}}{h^{2}}

4

λ_{0} = λ

Explanation:

A According to de-Brogile equation
$λ = \frac{h}{p}$
$and \sqrt{2 mE}$
$λ = \frac{h}{\sqrt{2 mE}}$
$E = \frac{hc}{λ_{o}}$
$λ = \frac{h}{\sqrt{2 m \times \frac{hc}{λ_{o}}}}$
$λ = \frac{2 mc λ^{2}}{h}$

JIPMER-2015

Dual nature of radiation and Matter

142561 If the kinetic energy of the particle is increased to 16 times its previous value, the percentage change in the de-Broglie wavelength of the particle is

1 25

2 75

3 60

4 50

Explanation:

B According to de-Broglie wavelength
$λ_{1} = \frac{h}{p} = \frac{h}{\sqrt{2 m (KE)}}$
$λ_{2} = \frac{h}{\sqrt{2 m 16 (KE)}} = \frac{h}{4 \sqrt{2 m (KE)}} = \frac{λ_{1}}{4}$
Change in wavelength $= λ_{1} - λ_{2}$
$= \frac{3 λ_{1}}{4}$
$% Change = \frac{\frac{3 λ_{1}}{4}}{λ_{1}} \times 100$
$= 75 %$

AIPMT - 2014

Dual nature of radiation and Matter

142563 The energy of a photon of light is $3 eV$ . Then the wavelength of photon must be

1

4125 nm

2

412.5 nm

3

41250 nm

4

4 nm

Explanation:

B Given,
Speed of light $(c) = 3 \times 10^{8} m / s$
Planck's constant $(h) = 6.6 \times 10^{- 34} JS$
$1 ev = 1.6 \times 10^{- 19} J$
$E = h v = 3 eV$
$\frac{hc}{λ} = 3 eV$
$λ = \frac{hc}{3 eV}$
$λ = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{3 \times 1.6 \times 10^{- 19}}$
$λ = \frac{19.8 \times 10^{- 26}}{4.8 \times 10^{- 19}}$
$λ = 412.5 nm$

AIPMT - 2000

Dual nature of radiation and Matter

142566 A $200 W$ sodium street lamp emits yellow light of wavelength $0.6 μ m$ . Assuming it to be $25 %$ efficient in converting electrical energy to light, the number of photons of yellow light it emits per second is

1

1.5 \times 10^{20}

2

6 \times 10^{18}

3

62 \times 10^{20}

4

3 \times 10^{19}

Explanation:

A Given,
Power $(P) = 200 W, λ = 0.6 μ m = 0.6 \times 10^{- 6} m$
Effective power $(E) = 200 \times \frac{25}{100}$
$= 50 W$
Number of photon per sec.
$\frac{n}{t} = \frac{P λ}{hc}$
$= \frac{50 \times 0.6 \times 10^{- 6}}{6.6 \times 10^{- 34} \times 3 \times 10^{8}}$
$\frac{n}{t} = 1.5 \times 10^{20}$ photons per secss

AIPMT - 2012

Dual nature of radiation and Matter

142560 Electrons of mass $m$ with de-Broglie wavelength
$λ$ fall on the target in an X-ray tube. The cutoff wavelength $(λ_{0})$ of the emitted $X$ -rays is

1

λ_{0} = \frac{2 mc λ^{2}}{h}

2

λ_{0} = \frac{2 h}{mc}

3

λ_{0} = \frac{2 m^{2} c^{2} λ^{3}}{h^{2}}

4

λ_{0} = λ

Explanation:

A According to de-Brogile equation
$λ = \frac{h}{p}$
$and \sqrt{2 mE}$
$λ = \frac{h}{\sqrt{2 mE}}$
$E = \frac{hc}{λ_{o}}$
$λ = \frac{h}{\sqrt{2 m \times \frac{hc}{λ_{o}}}}$
$λ = \frac{2 mc λ^{2}}{h}$

JIPMER-2015

Dual nature of radiation and Matter

142561 If the kinetic energy of the particle is increased to 16 times its previous value, the percentage change in the de-Broglie wavelength of the particle is

1 25

2 75

3 60

4 50

Explanation:

B According to de-Broglie wavelength
$λ_{1} = \frac{h}{p} = \frac{h}{\sqrt{2 m (KE)}}$
$λ_{2} = \frac{h}{\sqrt{2 m 16 (KE)}} = \frac{h}{4 \sqrt{2 m (KE)}} = \frac{λ_{1}}{4}$
Change in wavelength $= λ_{1} - λ_{2}$
$= \frac{3 λ_{1}}{4}$
$% Change = \frac{\frac{3 λ_{1}}{4}}{λ_{1}} \times 100$
$= 75 %$

AIPMT - 2014

Dual nature of radiation and Matter

142563 The energy of a photon of light is $3 eV$ . Then the wavelength of photon must be

1

4125 nm

2

412.5 nm

3

41250 nm

4

4 nm

Explanation:

B Given,
Speed of light $(c) = 3 \times 10^{8} m / s$
Planck's constant $(h) = 6.6 \times 10^{- 34} JS$
$1 ev = 1.6 \times 10^{- 19} J$
$E = h v = 3 eV$
$\frac{hc}{λ} = 3 eV$
$λ = \frac{hc}{3 eV}$
$λ = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{3 \times 1.6 \times 10^{- 19}}$
$λ = \frac{19.8 \times 10^{- 26}}{4.8 \times 10^{- 19}}$
$λ = 412.5 nm$

AIPMT - 2000

Dual nature of radiation and Matter

142566 A $200 W$ sodium street lamp emits yellow light of wavelength $0.6 μ m$ . Assuming it to be $25 %$ efficient in converting electrical energy to light, the number of photons of yellow light it emits per second is

1

1.5 \times 10^{20}

2

6 \times 10^{18}

3

62 \times 10^{20}

4

3 \times 10^{19}

Explanation:

A Given,
Power $(P) = 200 W, λ = 0.6 μ m = 0.6 \times 10^{- 6} m$
Effective power $(E) = 200 \times \frac{25}{100}$
$= 50 W$
Number of photon per sec.
$\frac{n}{t} = \frac{P λ}{hc}$
$= \frac{50 \times 0.6 \times 10^{- 6}}{6.6 \times 10^{- 34} \times 3 \times 10^{8}}$
$\frac{n}{t} = 1.5 \times 10^{20}$ photons per secss

AIPMT - 2012

Dual nature of radiation and Matter

142560 Electrons of mass $m$ with de-Broglie wavelength
$λ$ fall on the target in an X-ray tube. The cutoff wavelength $(λ_{0})$ of the emitted $X$ -rays is

1

λ_{0} = \frac{2 mc λ^{2}}{h}

2

λ_{0} = \frac{2 h}{mc}

3

λ_{0} = \frac{2 m^{2} c^{2} λ^{3}}{h^{2}}

4

λ_{0} = λ

Explanation:

A According to de-Brogile equation
$λ = \frac{h}{p}$
$and \sqrt{2 mE}$
$λ = \frac{h}{\sqrt{2 mE}}$
$E = \frac{hc}{λ_{o}}$
$λ = \frac{h}{\sqrt{2 m \times \frac{hc}{λ_{o}}}}$
$λ = \frac{2 mc λ^{2}}{h}$

JIPMER-2015

Dual nature of radiation and Matter

142561 If the kinetic energy of the particle is increased to 16 times its previous value, the percentage change in the de-Broglie wavelength of the particle is

1 25

2 75

3 60

4 50

Explanation:

B According to de-Broglie wavelength
$λ_{1} = \frac{h}{p} = \frac{h}{\sqrt{2 m (KE)}}$
$λ_{2} = \frac{h}{\sqrt{2 m 16 (KE)}} = \frac{h}{4 \sqrt{2 m (KE)}} = \frac{λ_{1}}{4}$
Change in wavelength $= λ_{1} - λ_{2}$
$= \frac{3 λ_{1}}{4}$
$% Change = \frac{\frac{3 λ_{1}}{4}}{λ_{1}} \times 100$
$= 75 %$

AIPMT - 2014

Dual nature of radiation and Matter

142563 The energy of a photon of light is $3 eV$ . Then the wavelength of photon must be

1

4125 nm

2

412.5 nm

3

41250 nm

4

4 nm

Explanation:

B Given,
Speed of light $(c) = 3 \times 10^{8} m / s$
Planck's constant $(h) = 6.6 \times 10^{- 34} JS$
$1 ev = 1.6 \times 10^{- 19} J$
$E = h v = 3 eV$
$\frac{hc}{λ} = 3 eV$
$λ = \frac{hc}{3 eV}$
$λ = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{3 \times 1.6 \times 10^{- 19}}$
$λ = \frac{19.8 \times 10^{- 26}}{4.8 \times 10^{- 19}}$
$λ = 412.5 nm$

AIPMT - 2000

Dual nature of radiation and Matter

142566 A $200 W$ sodium street lamp emits yellow light of wavelength $0.6 μ m$ . Assuming it to be $25 %$ efficient in converting electrical energy to light, the number of photons of yellow light it emits per second is

1

1.5 \times 10^{20}

2

6 \times 10^{18}

3

62 \times 10^{20}

4

3 \times 10^{19}

Explanation:

A Given,
Power $(P) = 200 W, λ = 0.6 μ m = 0.6 \times 10^{- 6} m$
Effective power $(E) = 200 \times \frac{25}{100}$
$= 50 W$
Number of photon per sec.
$\frac{n}{t} = \frac{P λ}{hc}$
$= \frac{50 \times 0.6 \times 10^{- 6}}{6.6 \times 10^{- 34} \times 3 \times 10^{8}}$
$\frac{n}{t} = 1.5 \times 10^{20}$ photons per secss

AIPMT - 2012

Dual nature of radiation and Matter

142560 Electrons of mass $m$ with de-Broglie wavelength
$λ$ fall on the target in an X-ray tube. The cutoff wavelength $(λ_{0})$ of the emitted $X$ -rays is

1

λ_{0} = \frac{2 mc λ^{2}}{h}

2

λ_{0} = \frac{2 h}{mc}

3

λ_{0} = \frac{2 m^{2} c^{2} λ^{3}}{h^{2}}

4

λ_{0} = λ

Explanation:

A According to de-Brogile equation
$λ = \frac{h}{p}$
$and \sqrt{2 mE}$
$λ = \frac{h}{\sqrt{2 mE}}$
$E = \frac{hc}{λ_{o}}$
$λ = \frac{h}{\sqrt{2 m \times \frac{hc}{λ_{o}}}}$
$λ = \frac{2 mc λ^{2}}{h}$

JIPMER-2015

Dual nature of radiation and Matter

142561 If the kinetic energy of the particle is increased to 16 times its previous value, the percentage change in the de-Broglie wavelength of the particle is

1 25

2 75

3 60

4 50

Explanation:

B According to de-Broglie wavelength
$λ_{1} = \frac{h}{p} = \frac{h}{\sqrt{2 m (KE)}}$
$λ_{2} = \frac{h}{\sqrt{2 m 16 (KE)}} = \frac{h}{4 \sqrt{2 m (KE)}} = \frac{λ_{1}}{4}$
Change in wavelength $= λ_{1} - λ_{2}$
$= \frac{3 λ_{1}}{4}$
$% Change = \frac{\frac{3 λ_{1}}{4}}{λ_{1}} \times 100$
$= 75 %$

AIPMT - 2014

Dual nature of radiation and Matter

142563 The energy of a photon of light is $3 eV$ . Then the wavelength of photon must be

1

4125 nm

2

412.5 nm

3

41250 nm

4

4 nm

Explanation:

B Given,
Speed of light $(c) = 3 \times 10^{8} m / s$
Planck's constant $(h) = 6.6 \times 10^{- 34} JS$
$1 ev = 1.6 \times 10^{- 19} J$
$E = h v = 3 eV$
$\frac{hc}{λ} = 3 eV$
$λ = \frac{hc}{3 eV}$
$λ = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{3 \times 1.6 \times 10^{- 19}}$
$λ = \frac{19.8 \times 10^{- 26}}{4.8 \times 10^{- 19}}$
$λ = 412.5 nm$

AIPMT - 2000

Dual nature of radiation and Matter

142566 A $200 W$ sodium street lamp emits yellow light of wavelength $0.6 μ m$ . Assuming it to be $25 %$ efficient in converting electrical energy to light, the number of photons of yellow light it emits per second is

1

1.5 \times 10^{20}

2

6 \times 10^{18}

3

62 \times 10^{20}

4

3 \times 10^{19}

Explanation:

A Given,
Power $(P) = 200 W, λ = 0.6 μ m = 0.6 \times 10^{- 6} m$
Effective power $(E) = 200 \times \frac{25}{100}$
$= 50 W$
Number of photon per sec.
$\frac{n}{t} = \frac{P λ}{hc}$
$= \frac{50 \times 0.6 \times 10^{- 6}}{6.6 \times 10^{- 34} \times 3 \times 10^{8}}$
$\frac{n}{t} = 1.5 \times 10^{20}$ photons per secss

AIPMT - 2012

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation: