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Structure of Atom

238714 If the kinetic energy of a particle is reduced to half, de-Broglie wavelength becomes :

1 2 times

2

\frac{1}{\sqrt{2}}

times

3 4 times

4

\sqrt{2}

times

Explanation:

: Given that, (K.E. $)_{2} = 1 / 2$ (K.E. $)_{1}$
According to de-Broglie equation
$λ = \frac{h}{mv} = \frac{h}{p} = \frac{h}{\sqrt{2 mK . E .}}$
Where, K.E. = Kinetic energy
So, the de-Broglie wavelength is inversely proportional to square root of kinetic energy (K.E.)
Therefore, $\frac{λ_{1}}{λ_{2}} = \sqrt{\frac{(K . E .)_{2}}{(K . E .)_{1}}}$
Then, $\frac{λ_{1}}{λ_{2}} = \sqrt{\frac{1}{2} \frac{(K . E .)_{1}}{(K . E .)_{1}}}$
or $λ_{2} = \sqrt{2} λ_{1}$
Thus, de-Broglie wavelength becomes $\sqrt{2}$ times to the original wavelength.

AP-EAMCET (Engg.) 2015

Structure of Atom

238681 The energy of mole of photons of radiation of wavelength $300 nm$ is (given $h = 6.63 \times 10^{- 34} Js$ , $N_{A} = 6.02 \times 10^{23} {mol}^{- 1} C = 3 \times 10^{8} {ms}^{- 1}$ )

1

235 kJ {mol}^{- 1}

2

325 kJ {mol}^{- 1}

3

399 kJ {mol}^{- 1}

4

435 kJ {mol}^{- 1}

Explanation:

: Given,
Wavelength of photon $= 300 nm$
Planck's constant $= 6.63 \times 10^{- 34} JS$
$\begin{matrix} N_{A} = 6.02 \times 10^{23} {mol}^{- 1} \\ C = 3 \times 10^{8} {ms}^{- 1} \end{matrix}$
As we know that-
$\begin{aligned} Energy = & \frac{hc}{λ} \\ Energy = & \frac{6.63 \times 10^{- 34} \times 3 \times 10^{8}}{300 \times 10^{- 9}} \\ = 6.63 \times 10^{- 19} J \end{aligned}$
Energy of one mole of photon
$\begin{aligned} = 6.63 \times 10^{- 19} \times 6.02 \times 10^{23} {mol}^{- 1} \\ = 399 kJ / mol . \end{aligned}$

JEE Main-24.06.2022

Structure of Atom

238682 If wavelength of photon is $2.2 \times 10^{- 11} m$ and $h = 6.6 \times 10^{- 34} J s$ , then momentum of photon

1

1.452 \times 10^{- 44} kg {ms}^{- 1}

2

6.89 \times 10^{43} kg {ms}^{- 1}

3

3 \times 10^{- 23} kg {ms}^{- 1}

4

3.33 \times 10^{- 22} kg {ms}^{- 1}

Explanation:

: Given that -
Wavelength of proton $(λ) = 2.2 \times 10^{- 11} m$
Planck's constant $= 6.6 \times 10^{- 34} JS$ .
Momentum of photon $=$ ?
According to de - Broglie equation
$\begin{aligned} λ = \frac{h}{p} \\ p = \frac{h}{λ} \\ p = \frac{6.6 \times 10^{- 34}}{2.2 \times 10^{- 11}} \\ p = 3 \times 10^{- 34} \times 10^{11} \\ p = 3 \times 10^{- 23} kg . {ms}^{- 1} \end{aligned}$

Karnataka CET-17.06.2022

Structure of Atom

238683 The minimum energy that must be possessed by photons in order to produce the photoelectric effect with platinum metal is [Given: The threshold frequency of platinum is $1.3 \times 10^{15} s^{- 1}$ and $h = 6.6 \times 10^{- 34} Js$ .]

1

3.21 \times 10^{- 14} J

2

6.24 \times 10^{- 16} J

3

8.58 \times 10^{- 19} J

4

9.76 \times 10^{- 20} J

Explanation:

: Given,
Threshold frequency of platinum $= 1.3 \times 10^{15} s^{- 1}$
Planck's constant $(h) = 6.6 \times 10^{- 34} JS$
As we know that-
$Energy = \frac{hc}{λ}$
And frequency $(v) = \frac{c}{λ}$
Then,
Energy $= h v$
$= 6.6 \times 10^{- 34} JS \times 1.3 \times 10^{15} s^{- 1}$
$= 8.58 \times 10^{- 19} J$

JEE Main-25.06.2022

Structure of Atom

238684 A particle of mass $6.6 \times 10^{- 31} kg$ is moving with a velocity of $1 \times 10^{7} {ms}^{- 1}$ . The de Broglie wavelength (in $\AA$ ) associated with the particle, is $(h = 6.6 \times 10^{- 34} Js)$

1 1

2 10

3 5

4 2

5 4

Explanation:

: Given that -
Mass particle $(m) = 6.6 \times 10^{- 31} kg$
Velocity of particle $(v) = 1 \times 10^{7} m / s$
Wave length $(λ) =$ ?
$h = 6.6 \times 10^{- 34} Js$
We know that -
$\begin{aligned} λ = \frac{h}{mv} \\ λ = \frac{6.6 \times 10^{- 34}}{6.6 \times 10^{- 31} \times 10^{7}} \\ λ = 10^{- 34} \times 10^{24} \\ λ = 10^{- 10} \\ λ = 1 A^{\circ} \\ ∴ 1 A^{\circ} = 10^{- 10} \end{aligned}$

Kerala CEE -03.07.2022

Structure of Atom

238714 If the kinetic energy of a particle is reduced to half, de-Broglie wavelength becomes :

1 2 times

2

\frac{1}{\sqrt{2}}

times

3 4 times

4

\sqrt{2}

times

Explanation:

: Given that, (K.E. $)_{2} = 1 / 2$ (K.E. $)_{1}$
According to de-Broglie equation
$λ = \frac{h}{mv} = \frac{h}{p} = \frac{h}{\sqrt{2 mK . E .}}$
Where, K.E. = Kinetic energy
So, the de-Broglie wavelength is inversely proportional to square root of kinetic energy (K.E.)
Therefore, $\frac{λ_{1}}{λ_{2}} = \sqrt{\frac{(K . E .)_{2}}{(K . E .)_{1}}}$
Then, $\frac{λ_{1}}{λ_{2}} = \sqrt{\frac{1}{2} \frac{(K . E .)_{1}}{(K . E .)_{1}}}$
or $λ_{2} = \sqrt{2} λ_{1}$
Thus, de-Broglie wavelength becomes $\sqrt{2}$ times to the original wavelength.

AP-EAMCET (Engg.) 2015

Structure of Atom

238681 The energy of mole of photons of radiation of wavelength $300 nm$ is (given $h = 6.63 \times 10^{- 34} Js$ , $N_{A} = 6.02 \times 10^{23} {mol}^{- 1} C = 3 \times 10^{8} {ms}^{- 1}$ )

1

235 kJ {mol}^{- 1}

2

325 kJ {mol}^{- 1}

3

399 kJ {mol}^{- 1}

4

435 kJ {mol}^{- 1}

Explanation:

: Given,
Wavelength of photon $= 300 nm$
Planck's constant $= 6.63 \times 10^{- 34} JS$
$\begin{matrix} N_{A} = 6.02 \times 10^{23} {mol}^{- 1} \\ C = 3 \times 10^{8} {ms}^{- 1} \end{matrix}$
As we know that-
$\begin{aligned} Energy = & \frac{hc}{λ} \\ Energy = & \frac{6.63 \times 10^{- 34} \times 3 \times 10^{8}}{300 \times 10^{- 9}} \\ = 6.63 \times 10^{- 19} J \end{aligned}$
Energy of one mole of photon
$\begin{aligned} = 6.63 \times 10^{- 19} \times 6.02 \times 10^{23} {mol}^{- 1} \\ = 399 kJ / mol . \end{aligned}$

JEE Main-24.06.2022

Structure of Atom

238682 If wavelength of photon is $2.2 \times 10^{- 11} m$ and $h = 6.6 \times 10^{- 34} J s$ , then momentum of photon

1

1.452 \times 10^{- 44} kg {ms}^{- 1}

2

6.89 \times 10^{43} kg {ms}^{- 1}

3

3 \times 10^{- 23} kg {ms}^{- 1}

4

3.33 \times 10^{- 22} kg {ms}^{- 1}

Explanation:

: Given that -
Wavelength of proton $(λ) = 2.2 \times 10^{- 11} m$
Planck's constant $= 6.6 \times 10^{- 34} JS$ .
Momentum of photon $=$ ?
According to de - Broglie equation
$\begin{aligned} λ = \frac{h}{p} \\ p = \frac{h}{λ} \\ p = \frac{6.6 \times 10^{- 34}}{2.2 \times 10^{- 11}} \\ p = 3 \times 10^{- 34} \times 10^{11} \\ p = 3 \times 10^{- 23} kg . {ms}^{- 1} \end{aligned}$

Karnataka CET-17.06.2022

Structure of Atom

238683 The minimum energy that must be possessed by photons in order to produce the photoelectric effect with platinum metal is [Given: The threshold frequency of platinum is $1.3 \times 10^{15} s^{- 1}$ and $h = 6.6 \times 10^{- 34} Js$ .]

1

3.21 \times 10^{- 14} J

2

6.24 \times 10^{- 16} J

3

8.58 \times 10^{- 19} J

4

9.76 \times 10^{- 20} J

Explanation:

: Given,
Threshold frequency of platinum $= 1.3 \times 10^{15} s^{- 1}$
Planck's constant $(h) = 6.6 \times 10^{- 34} JS$
As we know that-
$Energy = \frac{hc}{λ}$
And frequency $(v) = \frac{c}{λ}$
Then,
Energy $= h v$
$= 6.6 \times 10^{- 34} JS \times 1.3 \times 10^{15} s^{- 1}$
$= 8.58 \times 10^{- 19} J$

JEE Main-25.06.2022

Structure of Atom

238684 A particle of mass $6.6 \times 10^{- 31} kg$ is moving with a velocity of $1 \times 10^{7} {ms}^{- 1}$ . The de Broglie wavelength (in $\AA$ ) associated with the particle, is $(h = 6.6 \times 10^{- 34} Js)$

1 1

2 10

3 5

4 2

5 4

Explanation:

: Given that -
Mass particle $(m) = 6.6 \times 10^{- 31} kg$
Velocity of particle $(v) = 1 \times 10^{7} m / s$
Wave length $(λ) =$ ?
$h = 6.6 \times 10^{- 34} Js$
We know that -
$\begin{aligned} λ = \frac{h}{mv} \\ λ = \frac{6.6 \times 10^{- 34}}{6.6 \times 10^{- 31} \times 10^{7}} \\ λ = 10^{- 34} \times 10^{24} \\ λ = 10^{- 10} \\ λ = 1 A^{\circ} \\ ∴ 1 A^{\circ} = 10^{- 10} \end{aligned}$

Kerala CEE -03.07.2022

Structure of Atom

238714 If the kinetic energy of a particle is reduced to half, de-Broglie wavelength becomes :

1 2 times

2

\frac{1}{\sqrt{2}}

times

3 4 times

4

\sqrt{2}

times

Explanation:

: Given that, (K.E. $)_{2} = 1 / 2$ (K.E. $)_{1}$
According to de-Broglie equation
$λ = \frac{h}{mv} = \frac{h}{p} = \frac{h}{\sqrt{2 mK . E .}}$
Where, K.E. = Kinetic energy
So, the de-Broglie wavelength is inversely proportional to square root of kinetic energy (K.E.)
Therefore, $\frac{λ_{1}}{λ_{2}} = \sqrt{\frac{(K . E .)_{2}}{(K . E .)_{1}}}$
Then, $\frac{λ_{1}}{λ_{2}} = \sqrt{\frac{1}{2} \frac{(K . E .)_{1}}{(K . E .)_{1}}}$
or $λ_{2} = \sqrt{2} λ_{1}$
Thus, de-Broglie wavelength becomes $\sqrt{2}$ times to the original wavelength.

AP-EAMCET (Engg.) 2015

Structure of Atom

238681 The energy of mole of photons of radiation of wavelength $300 nm$ is (given $h = 6.63 \times 10^{- 34} Js$ , $N_{A} = 6.02 \times 10^{23} {mol}^{- 1} C = 3 \times 10^{8} {ms}^{- 1}$ )

1

235 kJ {mol}^{- 1}

2

325 kJ {mol}^{- 1}

3

399 kJ {mol}^{- 1}

4

435 kJ {mol}^{- 1}

Explanation:

: Given,
Wavelength of photon $= 300 nm$
Planck's constant $= 6.63 \times 10^{- 34} JS$
$\begin{matrix} N_{A} = 6.02 \times 10^{23} {mol}^{- 1} \\ C = 3 \times 10^{8} {ms}^{- 1} \end{matrix}$
As we know that-
$\begin{aligned} Energy = & \frac{hc}{λ} \\ Energy = & \frac{6.63 \times 10^{- 34} \times 3 \times 10^{8}}{300 \times 10^{- 9}} \\ = 6.63 \times 10^{- 19} J \end{aligned}$
Energy of one mole of photon
$\begin{aligned} = 6.63 \times 10^{- 19} \times 6.02 \times 10^{23} {mol}^{- 1} \\ = 399 kJ / mol . \end{aligned}$

JEE Main-24.06.2022

Structure of Atom

238682 If wavelength of photon is $2.2 \times 10^{- 11} m$ and $h = 6.6 \times 10^{- 34} J s$ , then momentum of photon

1

1.452 \times 10^{- 44} kg {ms}^{- 1}

2

6.89 \times 10^{43} kg {ms}^{- 1}

3

3 \times 10^{- 23} kg {ms}^{- 1}

4

3.33 \times 10^{- 22} kg {ms}^{- 1}

Explanation:

: Given that -
Wavelength of proton $(λ) = 2.2 \times 10^{- 11} m$
Planck's constant $= 6.6 \times 10^{- 34} JS$ .
Momentum of photon $=$ ?
According to de - Broglie equation
$\begin{aligned} λ = \frac{h}{p} \\ p = \frac{h}{λ} \\ p = \frac{6.6 \times 10^{- 34}}{2.2 \times 10^{- 11}} \\ p = 3 \times 10^{- 34} \times 10^{11} \\ p = 3 \times 10^{- 23} kg . {ms}^{- 1} \end{aligned}$

Karnataka CET-17.06.2022

Structure of Atom

238683 The minimum energy that must be possessed by photons in order to produce the photoelectric effect with platinum metal is [Given: The threshold frequency of platinum is $1.3 \times 10^{15} s^{- 1}$ and $h = 6.6 \times 10^{- 34} Js$ .]

1

3.21 \times 10^{- 14} J

2

6.24 \times 10^{- 16} J

3

8.58 \times 10^{- 19} J

4

9.76 \times 10^{- 20} J

Explanation:

: Given,
Threshold frequency of platinum $= 1.3 \times 10^{15} s^{- 1}$
Planck's constant $(h) = 6.6 \times 10^{- 34} JS$
As we know that-
$Energy = \frac{hc}{λ}$
And frequency $(v) = \frac{c}{λ}$
Then,
Energy $= h v$
$= 6.6 \times 10^{- 34} JS \times 1.3 \times 10^{15} s^{- 1}$
$= 8.58 \times 10^{- 19} J$

JEE Main-25.06.2022

Structure of Atom

238684 A particle of mass $6.6 \times 10^{- 31} kg$ is moving with a velocity of $1 \times 10^{7} {ms}^{- 1}$ . The de Broglie wavelength (in $\AA$ ) associated with the particle, is $(h = 6.6 \times 10^{- 34} Js)$

1 1

2 10

3 5

4 2

5 4

Explanation:

: Given that -
Mass particle $(m) = 6.6 \times 10^{- 31} kg$
Velocity of particle $(v) = 1 \times 10^{7} m / s$
Wave length $(λ) =$ ?
$h = 6.6 \times 10^{- 34} Js$
We know that -
$\begin{aligned} λ = \frac{h}{mv} \\ λ = \frac{6.6 \times 10^{- 34}}{6.6 \times 10^{- 31} \times 10^{7}} \\ λ = 10^{- 34} \times 10^{24} \\ λ = 10^{- 10} \\ λ = 1 A^{\circ} \\ ∴ 1 A^{\circ} = 10^{- 10} \end{aligned}$

Kerala CEE -03.07.2022

Structure of Atom

238714 If the kinetic energy of a particle is reduced to half, de-Broglie wavelength becomes :

1 2 times

2

\frac{1}{\sqrt{2}}

times

3 4 times

4

\sqrt{2}

times

Explanation:

: Given that, (K.E. $)_{2} = 1 / 2$ (K.E. $)_{1}$
According to de-Broglie equation
$λ = \frac{h}{mv} = \frac{h}{p} = \frac{h}{\sqrt{2 mK . E .}}$
Where, K.E. = Kinetic energy
So, the de-Broglie wavelength is inversely proportional to square root of kinetic energy (K.E.)
Therefore, $\frac{λ_{1}}{λ_{2}} = \sqrt{\frac{(K . E .)_{2}}{(K . E .)_{1}}}$
Then, $\frac{λ_{1}}{λ_{2}} = \sqrt{\frac{1}{2} \frac{(K . E .)_{1}}{(K . E .)_{1}}}$
or $λ_{2} = \sqrt{2} λ_{1}$
Thus, de-Broglie wavelength becomes $\sqrt{2}$ times to the original wavelength.

AP-EAMCET (Engg.) 2015

Structure of Atom

238681 The energy of mole of photons of radiation of wavelength $300 nm$ is (given $h = 6.63 \times 10^{- 34} Js$ , $N_{A} = 6.02 \times 10^{23} {mol}^{- 1} C = 3 \times 10^{8} {ms}^{- 1}$ )

1

235 kJ {mol}^{- 1}

2

325 kJ {mol}^{- 1}

3

399 kJ {mol}^{- 1}

4

435 kJ {mol}^{- 1}

Explanation:

: Given,
Wavelength of photon $= 300 nm$
Planck's constant $= 6.63 \times 10^{- 34} JS$
$\begin{matrix} N_{A} = 6.02 \times 10^{23} {mol}^{- 1} \\ C = 3 \times 10^{8} {ms}^{- 1} \end{matrix}$
As we know that-
$\begin{aligned} Energy = & \frac{hc}{λ} \\ Energy = & \frac{6.63 \times 10^{- 34} \times 3 \times 10^{8}}{300 \times 10^{- 9}} \\ = 6.63 \times 10^{- 19} J \end{aligned}$
Energy of one mole of photon
$\begin{aligned} = 6.63 \times 10^{- 19} \times 6.02 \times 10^{23} {mol}^{- 1} \\ = 399 kJ / mol . \end{aligned}$

JEE Main-24.06.2022

Structure of Atom

238682 If wavelength of photon is $2.2 \times 10^{- 11} m$ and $h = 6.6 \times 10^{- 34} J s$ , then momentum of photon

1

1.452 \times 10^{- 44} kg {ms}^{- 1}

2

6.89 \times 10^{43} kg {ms}^{- 1}

3

3 \times 10^{- 23} kg {ms}^{- 1}

4

3.33 \times 10^{- 22} kg {ms}^{- 1}

Explanation:

: Given that -
Wavelength of proton $(λ) = 2.2 \times 10^{- 11} m$
Planck's constant $= 6.6 \times 10^{- 34} JS$ .
Momentum of photon $=$ ?
According to de - Broglie equation
$\begin{aligned} λ = \frac{h}{p} \\ p = \frac{h}{λ} \\ p = \frac{6.6 \times 10^{- 34}}{2.2 \times 10^{- 11}} \\ p = 3 \times 10^{- 34} \times 10^{11} \\ p = 3 \times 10^{- 23} kg . {ms}^{- 1} \end{aligned}$

Karnataka CET-17.06.2022

Structure of Atom

238683 The minimum energy that must be possessed by photons in order to produce the photoelectric effect with platinum metal is [Given: The threshold frequency of platinum is $1.3 \times 10^{15} s^{- 1}$ and $h = 6.6 \times 10^{- 34} Js$ .]

1

3.21 \times 10^{- 14} J

2

6.24 \times 10^{- 16} J

3

8.58 \times 10^{- 19} J

4

9.76 \times 10^{- 20} J

Explanation:

: Given,
Threshold frequency of platinum $= 1.3 \times 10^{15} s^{- 1}$
Planck's constant $(h) = 6.6 \times 10^{- 34} JS$
As we know that-
$Energy = \frac{hc}{λ}$
And frequency $(v) = \frac{c}{λ}$
Then,
Energy $= h v$
$= 6.6 \times 10^{- 34} JS \times 1.3 \times 10^{15} s^{- 1}$
$= 8.58 \times 10^{- 19} J$

JEE Main-25.06.2022

Structure of Atom

238684 A particle of mass $6.6 \times 10^{- 31} kg$ is moving with a velocity of $1 \times 10^{7} {ms}^{- 1}$ . The de Broglie wavelength (in $\AA$ ) associated with the particle, is $(h = 6.6 \times 10^{- 34} Js)$

1 1

2 10

3 5

4 2

5 4

Explanation:

: Given that -
Mass particle $(m) = 6.6 \times 10^{- 31} kg$
Velocity of particle $(v) = 1 \times 10^{7} m / s$
Wave length $(λ) =$ ?
$h = 6.6 \times 10^{- 34} Js$
We know that -
$\begin{aligned} λ = \frac{h}{mv} \\ λ = \frac{6.6 \times 10^{- 34}}{6.6 \times 10^{- 31} \times 10^{7}} \\ λ = 10^{- 34} \times 10^{24} \\ λ = 10^{- 10} \\ λ = 1 A^{\circ} \\ ∴ 1 A^{\circ} = 10^{- 10} \end{aligned}$

Kerala CEE -03.07.2022

Structure of Atom

238714 If the kinetic energy of a particle is reduced to half, de-Broglie wavelength becomes :

1 2 times

2

\frac{1}{\sqrt{2}}

times

3 4 times

4

\sqrt{2}

times

Explanation:

: Given that, (K.E. $)_{2} = 1 / 2$ (K.E. $)_{1}$
According to de-Broglie equation
$λ = \frac{h}{mv} = \frac{h}{p} = \frac{h}{\sqrt{2 mK . E .}}$
Where, K.E. = Kinetic energy
So, the de-Broglie wavelength is inversely proportional to square root of kinetic energy (K.E.)
Therefore, $\frac{λ_{1}}{λ_{2}} = \sqrt{\frac{(K . E .)_{2}}{(K . E .)_{1}}}$
Then, $\frac{λ_{1}}{λ_{2}} = \sqrt{\frac{1}{2} \frac{(K . E .)_{1}}{(K . E .)_{1}}}$
or $λ_{2} = \sqrt{2} λ_{1}$
Thus, de-Broglie wavelength becomes $\sqrt{2}$ times to the original wavelength.

AP-EAMCET (Engg.) 2015

Structure of Atom

238681 The energy of mole of photons of radiation of wavelength $300 nm$ is (given $h = 6.63 \times 10^{- 34} Js$ , $N_{A} = 6.02 \times 10^{23} {mol}^{- 1} C = 3 \times 10^{8} {ms}^{- 1}$ )

1

235 kJ {mol}^{- 1}

2

325 kJ {mol}^{- 1}

3

399 kJ {mol}^{- 1}

4

435 kJ {mol}^{- 1}

Explanation:

: Given,
Wavelength of photon $= 300 nm$
Planck's constant $= 6.63 \times 10^{- 34} JS$
$\begin{matrix} N_{A} = 6.02 \times 10^{23} {mol}^{- 1} \\ C = 3 \times 10^{8} {ms}^{- 1} \end{matrix}$
As we know that-
$\begin{aligned} Energy = & \frac{hc}{λ} \\ Energy = & \frac{6.63 \times 10^{- 34} \times 3 \times 10^{8}}{300 \times 10^{- 9}} \\ = 6.63 \times 10^{- 19} J \end{aligned}$
Energy of one mole of photon
$\begin{aligned} = 6.63 \times 10^{- 19} \times 6.02 \times 10^{23} {mol}^{- 1} \\ = 399 kJ / mol . \end{aligned}$

JEE Main-24.06.2022

Structure of Atom

238682 If wavelength of photon is $2.2 \times 10^{- 11} m$ and $h = 6.6 \times 10^{- 34} J s$ , then momentum of photon

1

1.452 \times 10^{- 44} kg {ms}^{- 1}

2

6.89 \times 10^{43} kg {ms}^{- 1}

3

3 \times 10^{- 23} kg {ms}^{- 1}

4

3.33 \times 10^{- 22} kg {ms}^{- 1}

Explanation:

: Given that -
Wavelength of proton $(λ) = 2.2 \times 10^{- 11} m$
Planck's constant $= 6.6 \times 10^{- 34} JS$ .
Momentum of photon $=$ ?
According to de - Broglie equation
$\begin{aligned} λ = \frac{h}{p} \\ p = \frac{h}{λ} \\ p = \frac{6.6 \times 10^{- 34}}{2.2 \times 10^{- 11}} \\ p = 3 \times 10^{- 34} \times 10^{11} \\ p = 3 \times 10^{- 23} kg . {ms}^{- 1} \end{aligned}$

Karnataka CET-17.06.2022

Structure of Atom

238683 The minimum energy that must be possessed by photons in order to produce the photoelectric effect with platinum metal is [Given: The threshold frequency of platinum is $1.3 \times 10^{15} s^{- 1}$ and $h = 6.6 \times 10^{- 34} Js$ .]

1

3.21 \times 10^{- 14} J

2

6.24 \times 10^{- 16} J

3

8.58 \times 10^{- 19} J

4

9.76 \times 10^{- 20} J

Explanation:

: Given,
Threshold frequency of platinum $= 1.3 \times 10^{15} s^{- 1}$
Planck's constant $(h) = 6.6 \times 10^{- 34} JS$
As we know that-
$Energy = \frac{hc}{λ}$
And frequency $(v) = \frac{c}{λ}$
Then,
Energy $= h v$
$= 6.6 \times 10^{- 34} JS \times 1.3 \times 10^{15} s^{- 1}$
$= 8.58 \times 10^{- 19} J$

JEE Main-25.06.2022

Structure of Atom

238684 A particle of mass $6.6 \times 10^{- 31} kg$ is moving with a velocity of $1 \times 10^{7} {ms}^{- 1}$ . The de Broglie wavelength (in $\AA$ ) associated with the particle, is $(h = 6.6 \times 10^{- 34} Js)$

1 1

2 10

3 5

4 2

5 4

Explanation:

: Given that -
Mass particle $(m) = 6.6 \times 10^{- 31} kg$
Velocity of particle $(v) = 1 \times 10^{7} m / s$
Wave length $(λ) =$ ?
$h = 6.6 \times 10^{- 34} Js$
We know that -
$\begin{aligned} λ = \frac{h}{mv} \\ λ = \frac{6.6 \times 10^{- 34}}{6.6 \times 10^{- 31} \times 10^{7}} \\ λ = 10^{- 34} \times 10^{24} \\ λ = 10^{- 10} \\ λ = 1 A^{\circ} \\ ∴ 1 A^{\circ} = 10^{- 10} \end{aligned}$

Kerala CEE -03.07.2022