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 $\lambda=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK} . \mathrm{E} .}}$ Where, K.E. = Kinetic energy So, the de-Broglie wavelength is inversely proportional to square root of kinetic energy (K.E.) Therefore, $\frac{\lambda_1}{\lambda_2}=\sqrt{\frac{(\mathrm{K} . \mathrm{E} .)_2}{(\mathrm{~K} . \mathrm{E} .)_1}}$ Then, $\frac{\lambda_1}{\lambda_2}=\sqrt{\frac{1}{2} \frac{(\mathrm{K} . \mathrm{E} .)_1}{(\mathrm{~K} . \mathrm{E} .)_1}}$ or $\quad \lambda_2=\sqrt{2} \lambda_1$ 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 $300 \mathrm{~nm}$ is (given $h=6.63 \times 10^{-34} \mathrm{Js}$, $\mathrm{N}_{\mathrm{A}}=6.02 \times 10^{23} \mathrm{~mol}^{-1} \mathrm{C}=3 \times 10^8 \mathrm{~ms}^{-1}$ )
1 $235 \mathrm{~kJ} \mathrm{~mol}^{-1}$
2 $325 \mathrm{~kJ} \mathrm{~mol}^{-1}$
3 $399 \mathrm{~kJ} \mathrm{~mol}^{-1}$
4 $435 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Explanation:
: Given, Wavelength of photon $=300 \mathrm{~nm}$ Planck's constant $=6.63 \times 10^{-34} \mathrm{JS}$ $\begin{gathered} \mathrm{N}_{\mathrm{A}}=6.02 \times 10^{23} \mathrm{~mol}^{-1} \\ \mathrm{C}=3 \times 10^8 \mathrm{~ms}^{-1} \end{gathered}$ As we know that- $\begin{aligned} \text { Energy }= & \frac{\mathrm{hc}}{\lambda} \\ \text { Energy }= & \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{300 \times 10^{-9}} \\ & =6.63 \times 10^{-19} \mathrm{~J} \end{aligned}$ Energy of one mole of photon \begin{aligned} & =6.63 \times 10^{-19} \times 6.02 \times 10^{23} \mathrm{~mol}^{-1} \\ & =399 \mathrm{~kJ} / \mathrm{mol} . \end{aligned}
JEE Main-24.06.2022
Structure of Atom
238682
If wavelength of photon is $2.2 \times 10^{-11} \mathrm{~m}$ and $\mathrm{h}=6.6 \times 10^{-34} \mathrm{~J} \mathrm{~s}$, then momentum of photon
: Given that - Wavelength of proton $(\lambda)=2.2 \times 10^{-11} \mathrm{~m}$ Planck's constant $=6.6 \times 10^{-34} \mathrm{JS}$. Momentum of photon $=$ ? According to de - Broglie equation $\begin{aligned} & \lambda=\frac{\mathrm{h}}{\mathrm{p}} \\ & \mathrm{p}=\frac{\mathrm{h}}{\lambda} \\ & \mathrm{p}=\frac{6.6 \times 10^{-34}}{2.2 \times 10^{-11}} \\ & \mathrm{p}=3 \times 10^{-34} \times 10^{11} \\ & \mathrm{p}=3 \times 10^{-23} \mathrm{~kg} . \mathrm{ms}^{-1} \\ & \end{aligned}$
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 $1.3 \times 10^{15} \mathrm{~s}^{-1}$ and $\mathrm{h}=6.6 \times 10^{-34} \mathrm{Js}$.]
1 $3.21 \times 10^{-14} \mathrm{~J}$
2 $6.24 \times 10^{-16} \mathrm{~J}$
3 $8.58 \times 10^{-19} \mathrm{~J}$
4 $9.76 \times 10^{-20} \mathrm{~J}$
Explanation:
: Given, Threshold frequency of platinum $=1.3 \times 10^{15} \mathrm{~s}^{-1}$ Planck's constant $(\mathrm{h})=6.6 \times 10^{-34} \mathrm{JS}$ As we know that- $\text { Energy }=\frac{\mathrm{hc}}{\lambda}$ And frequency $(v)=\frac{\mathrm{c}}{\lambda}$ Then, Energy $=\mathrm{h} v$ $=6.6 \times 10^{-34} \mathrm{JS} \times 1.3 \times 10^{15} \mathrm{~s}^{-1}$ $=8.58 \times 10^{-19} \mathrm{~J}$
JEE Main-25.06.2022
Structure of Atom
238684
A particle of mass $6.6 \times 10^{-31} \mathrm{~kg}$ is moving with a velocity of $1 \times 10^7 \mathrm{~ms}^{-1}$. The de Broglie wavelength (in $\AA$ ) associated with the particle, is $\left(\mathrm{h}=6.6 \times 10^{-34} \mathrm{Js}\right)$
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 $\lambda=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK} . \mathrm{E} .}}$ Where, K.E. = Kinetic energy So, the de-Broglie wavelength is inversely proportional to square root of kinetic energy (K.E.) Therefore, $\frac{\lambda_1}{\lambda_2}=\sqrt{\frac{(\mathrm{K} . \mathrm{E} .)_2}{(\mathrm{~K} . \mathrm{E} .)_1}}$ Then, $\frac{\lambda_1}{\lambda_2}=\sqrt{\frac{1}{2} \frac{(\mathrm{K} . \mathrm{E} .)_1}{(\mathrm{~K} . \mathrm{E} .)_1}}$ or $\quad \lambda_2=\sqrt{2} \lambda_1$ 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 $300 \mathrm{~nm}$ is (given $h=6.63 \times 10^{-34} \mathrm{Js}$, $\mathrm{N}_{\mathrm{A}}=6.02 \times 10^{23} \mathrm{~mol}^{-1} \mathrm{C}=3 \times 10^8 \mathrm{~ms}^{-1}$ )
1 $235 \mathrm{~kJ} \mathrm{~mol}^{-1}$
2 $325 \mathrm{~kJ} \mathrm{~mol}^{-1}$
3 $399 \mathrm{~kJ} \mathrm{~mol}^{-1}$
4 $435 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Explanation:
: Given, Wavelength of photon $=300 \mathrm{~nm}$ Planck's constant $=6.63 \times 10^{-34} \mathrm{JS}$ $\begin{gathered} \mathrm{N}_{\mathrm{A}}=6.02 \times 10^{23} \mathrm{~mol}^{-1} \\ \mathrm{C}=3 \times 10^8 \mathrm{~ms}^{-1} \end{gathered}$ As we know that- $\begin{aligned} \text { Energy }= & \frac{\mathrm{hc}}{\lambda} \\ \text { Energy }= & \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{300 \times 10^{-9}} \\ & =6.63 \times 10^{-19} \mathrm{~J} \end{aligned}$ Energy of one mole of photon \begin{aligned} & =6.63 \times 10^{-19} \times 6.02 \times 10^{23} \mathrm{~mol}^{-1} \\ & =399 \mathrm{~kJ} / \mathrm{mol} . \end{aligned}
JEE Main-24.06.2022
Structure of Atom
238682
If wavelength of photon is $2.2 \times 10^{-11} \mathrm{~m}$ and $\mathrm{h}=6.6 \times 10^{-34} \mathrm{~J} \mathrm{~s}$, then momentum of photon
: Given that - Wavelength of proton $(\lambda)=2.2 \times 10^{-11} \mathrm{~m}$ Planck's constant $=6.6 \times 10^{-34} \mathrm{JS}$. Momentum of photon $=$ ? According to de - Broglie equation $\begin{aligned} & \lambda=\frac{\mathrm{h}}{\mathrm{p}} \\ & \mathrm{p}=\frac{\mathrm{h}}{\lambda} \\ & \mathrm{p}=\frac{6.6 \times 10^{-34}}{2.2 \times 10^{-11}} \\ & \mathrm{p}=3 \times 10^{-34} \times 10^{11} \\ & \mathrm{p}=3 \times 10^{-23} \mathrm{~kg} . \mathrm{ms}^{-1} \\ & \end{aligned}$
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 $1.3 \times 10^{15} \mathrm{~s}^{-1}$ and $\mathrm{h}=6.6 \times 10^{-34} \mathrm{Js}$.]
1 $3.21 \times 10^{-14} \mathrm{~J}$
2 $6.24 \times 10^{-16} \mathrm{~J}$
3 $8.58 \times 10^{-19} \mathrm{~J}$
4 $9.76 \times 10^{-20} \mathrm{~J}$
Explanation:
: Given, Threshold frequency of platinum $=1.3 \times 10^{15} \mathrm{~s}^{-1}$ Planck's constant $(\mathrm{h})=6.6 \times 10^{-34} \mathrm{JS}$ As we know that- $\text { Energy }=\frac{\mathrm{hc}}{\lambda}$ And frequency $(v)=\frac{\mathrm{c}}{\lambda}$ Then, Energy $=\mathrm{h} v$ $=6.6 \times 10^{-34} \mathrm{JS} \times 1.3 \times 10^{15} \mathrm{~s}^{-1}$ $=8.58 \times 10^{-19} \mathrm{~J}$
JEE Main-25.06.2022
Structure of Atom
238684
A particle of mass $6.6 \times 10^{-31} \mathrm{~kg}$ is moving with a velocity of $1 \times 10^7 \mathrm{~ms}^{-1}$. The de Broglie wavelength (in $\AA$ ) associated with the particle, is $\left(\mathrm{h}=6.6 \times 10^{-34} \mathrm{Js}\right)$
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 $\lambda=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK} . \mathrm{E} .}}$ Where, K.E. = Kinetic energy So, the de-Broglie wavelength is inversely proportional to square root of kinetic energy (K.E.) Therefore, $\frac{\lambda_1}{\lambda_2}=\sqrt{\frac{(\mathrm{K} . \mathrm{E} .)_2}{(\mathrm{~K} . \mathrm{E} .)_1}}$ Then, $\frac{\lambda_1}{\lambda_2}=\sqrt{\frac{1}{2} \frac{(\mathrm{K} . \mathrm{E} .)_1}{(\mathrm{~K} . \mathrm{E} .)_1}}$ or $\quad \lambda_2=\sqrt{2} \lambda_1$ 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 $300 \mathrm{~nm}$ is (given $h=6.63 \times 10^{-34} \mathrm{Js}$, $\mathrm{N}_{\mathrm{A}}=6.02 \times 10^{23} \mathrm{~mol}^{-1} \mathrm{C}=3 \times 10^8 \mathrm{~ms}^{-1}$ )
1 $235 \mathrm{~kJ} \mathrm{~mol}^{-1}$
2 $325 \mathrm{~kJ} \mathrm{~mol}^{-1}$
3 $399 \mathrm{~kJ} \mathrm{~mol}^{-1}$
4 $435 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Explanation:
: Given, Wavelength of photon $=300 \mathrm{~nm}$ Planck's constant $=6.63 \times 10^{-34} \mathrm{JS}$ $\begin{gathered} \mathrm{N}_{\mathrm{A}}=6.02 \times 10^{23} \mathrm{~mol}^{-1} \\ \mathrm{C}=3 \times 10^8 \mathrm{~ms}^{-1} \end{gathered}$ As we know that- $\begin{aligned} \text { Energy }= & \frac{\mathrm{hc}}{\lambda} \\ \text { Energy }= & \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{300 \times 10^{-9}} \\ & =6.63 \times 10^{-19} \mathrm{~J} \end{aligned}$ Energy of one mole of photon \begin{aligned} & =6.63 \times 10^{-19} \times 6.02 \times 10^{23} \mathrm{~mol}^{-1} \\ & =399 \mathrm{~kJ} / \mathrm{mol} . \end{aligned}
JEE Main-24.06.2022
Structure of Atom
238682
If wavelength of photon is $2.2 \times 10^{-11} \mathrm{~m}$ and $\mathrm{h}=6.6 \times 10^{-34} \mathrm{~J} \mathrm{~s}$, then momentum of photon
: Given that - Wavelength of proton $(\lambda)=2.2 \times 10^{-11} \mathrm{~m}$ Planck's constant $=6.6 \times 10^{-34} \mathrm{JS}$. Momentum of photon $=$ ? According to de - Broglie equation $\begin{aligned} & \lambda=\frac{\mathrm{h}}{\mathrm{p}} \\ & \mathrm{p}=\frac{\mathrm{h}}{\lambda} \\ & \mathrm{p}=\frac{6.6 \times 10^{-34}}{2.2 \times 10^{-11}} \\ & \mathrm{p}=3 \times 10^{-34} \times 10^{11} \\ & \mathrm{p}=3 \times 10^{-23} \mathrm{~kg} . \mathrm{ms}^{-1} \\ & \end{aligned}$
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 $1.3 \times 10^{15} \mathrm{~s}^{-1}$ and $\mathrm{h}=6.6 \times 10^{-34} \mathrm{Js}$.]
1 $3.21 \times 10^{-14} \mathrm{~J}$
2 $6.24 \times 10^{-16} \mathrm{~J}$
3 $8.58 \times 10^{-19} \mathrm{~J}$
4 $9.76 \times 10^{-20} \mathrm{~J}$
Explanation:
: Given, Threshold frequency of platinum $=1.3 \times 10^{15} \mathrm{~s}^{-1}$ Planck's constant $(\mathrm{h})=6.6 \times 10^{-34} \mathrm{JS}$ As we know that- $\text { Energy }=\frac{\mathrm{hc}}{\lambda}$ And frequency $(v)=\frac{\mathrm{c}}{\lambda}$ Then, Energy $=\mathrm{h} v$ $=6.6 \times 10^{-34} \mathrm{JS} \times 1.3 \times 10^{15} \mathrm{~s}^{-1}$ $=8.58 \times 10^{-19} \mathrm{~J}$
JEE Main-25.06.2022
Structure of Atom
238684
A particle of mass $6.6 \times 10^{-31} \mathrm{~kg}$ is moving with a velocity of $1 \times 10^7 \mathrm{~ms}^{-1}$. The de Broglie wavelength (in $\AA$ ) associated with the particle, is $\left(\mathrm{h}=6.6 \times 10^{-34} \mathrm{Js}\right)$
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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 $\lambda=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK} . \mathrm{E} .}}$ Where, K.E. = Kinetic energy So, the de-Broglie wavelength is inversely proportional to square root of kinetic energy (K.E.) Therefore, $\frac{\lambda_1}{\lambda_2}=\sqrt{\frac{(\mathrm{K} . \mathrm{E} .)_2}{(\mathrm{~K} . \mathrm{E} .)_1}}$ Then, $\frac{\lambda_1}{\lambda_2}=\sqrt{\frac{1}{2} \frac{(\mathrm{K} . \mathrm{E} .)_1}{(\mathrm{~K} . \mathrm{E} .)_1}}$ or $\quad \lambda_2=\sqrt{2} \lambda_1$ 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 $300 \mathrm{~nm}$ is (given $h=6.63 \times 10^{-34} \mathrm{Js}$, $\mathrm{N}_{\mathrm{A}}=6.02 \times 10^{23} \mathrm{~mol}^{-1} \mathrm{C}=3 \times 10^8 \mathrm{~ms}^{-1}$ )
1 $235 \mathrm{~kJ} \mathrm{~mol}^{-1}$
2 $325 \mathrm{~kJ} \mathrm{~mol}^{-1}$
3 $399 \mathrm{~kJ} \mathrm{~mol}^{-1}$
4 $435 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Explanation:
: Given, Wavelength of photon $=300 \mathrm{~nm}$ Planck's constant $=6.63 \times 10^{-34} \mathrm{JS}$ $\begin{gathered} \mathrm{N}_{\mathrm{A}}=6.02 \times 10^{23} \mathrm{~mol}^{-1} \\ \mathrm{C}=3 \times 10^8 \mathrm{~ms}^{-1} \end{gathered}$ As we know that- $\begin{aligned} \text { Energy }= & \frac{\mathrm{hc}}{\lambda} \\ \text { Energy }= & \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{300 \times 10^{-9}} \\ & =6.63 \times 10^{-19} \mathrm{~J} \end{aligned}$ Energy of one mole of photon \begin{aligned} & =6.63 \times 10^{-19} \times 6.02 \times 10^{23} \mathrm{~mol}^{-1} \\ & =399 \mathrm{~kJ} / \mathrm{mol} . \end{aligned}
JEE Main-24.06.2022
Structure of Atom
238682
If wavelength of photon is $2.2 \times 10^{-11} \mathrm{~m}$ and $\mathrm{h}=6.6 \times 10^{-34} \mathrm{~J} \mathrm{~s}$, then momentum of photon
: Given that - Wavelength of proton $(\lambda)=2.2 \times 10^{-11} \mathrm{~m}$ Planck's constant $=6.6 \times 10^{-34} \mathrm{JS}$. Momentum of photon $=$ ? According to de - Broglie equation $\begin{aligned} & \lambda=\frac{\mathrm{h}}{\mathrm{p}} \\ & \mathrm{p}=\frac{\mathrm{h}}{\lambda} \\ & \mathrm{p}=\frac{6.6 \times 10^{-34}}{2.2 \times 10^{-11}} \\ & \mathrm{p}=3 \times 10^{-34} \times 10^{11} \\ & \mathrm{p}=3 \times 10^{-23} \mathrm{~kg} . \mathrm{ms}^{-1} \\ & \end{aligned}$
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 $1.3 \times 10^{15} \mathrm{~s}^{-1}$ and $\mathrm{h}=6.6 \times 10^{-34} \mathrm{Js}$.]
1 $3.21 \times 10^{-14} \mathrm{~J}$
2 $6.24 \times 10^{-16} \mathrm{~J}$
3 $8.58 \times 10^{-19} \mathrm{~J}$
4 $9.76 \times 10^{-20} \mathrm{~J}$
Explanation:
: Given, Threshold frequency of platinum $=1.3 \times 10^{15} \mathrm{~s}^{-1}$ Planck's constant $(\mathrm{h})=6.6 \times 10^{-34} \mathrm{JS}$ As we know that- $\text { Energy }=\frac{\mathrm{hc}}{\lambda}$ And frequency $(v)=\frac{\mathrm{c}}{\lambda}$ Then, Energy $=\mathrm{h} v$ $=6.6 \times 10^{-34} \mathrm{JS} \times 1.3 \times 10^{15} \mathrm{~s}^{-1}$ $=8.58 \times 10^{-19} \mathrm{~J}$
JEE Main-25.06.2022
Structure of Atom
238684
A particle of mass $6.6 \times 10^{-31} \mathrm{~kg}$ is moving with a velocity of $1 \times 10^7 \mathrm{~ms}^{-1}$. The de Broglie wavelength (in $\AA$ ) associated with the particle, is $\left(\mathrm{h}=6.6 \times 10^{-34} \mathrm{Js}\right)$
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 $\lambda=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK} . \mathrm{E} .}}$ Where, K.E. = Kinetic energy So, the de-Broglie wavelength is inversely proportional to square root of kinetic energy (K.E.) Therefore, $\frac{\lambda_1}{\lambda_2}=\sqrt{\frac{(\mathrm{K} . \mathrm{E} .)_2}{(\mathrm{~K} . \mathrm{E} .)_1}}$ Then, $\frac{\lambda_1}{\lambda_2}=\sqrt{\frac{1}{2} \frac{(\mathrm{K} . \mathrm{E} .)_1}{(\mathrm{~K} . \mathrm{E} .)_1}}$ or $\quad \lambda_2=\sqrt{2} \lambda_1$ 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 $300 \mathrm{~nm}$ is (given $h=6.63 \times 10^{-34} \mathrm{Js}$, $\mathrm{N}_{\mathrm{A}}=6.02 \times 10^{23} \mathrm{~mol}^{-1} \mathrm{C}=3 \times 10^8 \mathrm{~ms}^{-1}$ )
1 $235 \mathrm{~kJ} \mathrm{~mol}^{-1}$
2 $325 \mathrm{~kJ} \mathrm{~mol}^{-1}$
3 $399 \mathrm{~kJ} \mathrm{~mol}^{-1}$
4 $435 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Explanation:
: Given, Wavelength of photon $=300 \mathrm{~nm}$ Planck's constant $=6.63 \times 10^{-34} \mathrm{JS}$ $\begin{gathered} \mathrm{N}_{\mathrm{A}}=6.02 \times 10^{23} \mathrm{~mol}^{-1} \\ \mathrm{C}=3 \times 10^8 \mathrm{~ms}^{-1} \end{gathered}$ As we know that- $\begin{aligned} \text { Energy }= & \frac{\mathrm{hc}}{\lambda} \\ \text { Energy }= & \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{300 \times 10^{-9}} \\ & =6.63 \times 10^{-19} \mathrm{~J} \end{aligned}$ Energy of one mole of photon \begin{aligned} & =6.63 \times 10^{-19} \times 6.02 \times 10^{23} \mathrm{~mol}^{-1} \\ & =399 \mathrm{~kJ} / \mathrm{mol} . \end{aligned}
JEE Main-24.06.2022
Structure of Atom
238682
If wavelength of photon is $2.2 \times 10^{-11} \mathrm{~m}$ and $\mathrm{h}=6.6 \times 10^{-34} \mathrm{~J} \mathrm{~s}$, then momentum of photon
: Given that - Wavelength of proton $(\lambda)=2.2 \times 10^{-11} \mathrm{~m}$ Planck's constant $=6.6 \times 10^{-34} \mathrm{JS}$. Momentum of photon $=$ ? According to de - Broglie equation $\begin{aligned} & \lambda=\frac{\mathrm{h}}{\mathrm{p}} \\ & \mathrm{p}=\frac{\mathrm{h}}{\lambda} \\ & \mathrm{p}=\frac{6.6 \times 10^{-34}}{2.2 \times 10^{-11}} \\ & \mathrm{p}=3 \times 10^{-34} \times 10^{11} \\ & \mathrm{p}=3 \times 10^{-23} \mathrm{~kg} . \mathrm{ms}^{-1} \\ & \end{aligned}$
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 $1.3 \times 10^{15} \mathrm{~s}^{-1}$ and $\mathrm{h}=6.6 \times 10^{-34} \mathrm{Js}$.]
1 $3.21 \times 10^{-14} \mathrm{~J}$
2 $6.24 \times 10^{-16} \mathrm{~J}$
3 $8.58 \times 10^{-19} \mathrm{~J}$
4 $9.76 \times 10^{-20} \mathrm{~J}$
Explanation:
: Given, Threshold frequency of platinum $=1.3 \times 10^{15} \mathrm{~s}^{-1}$ Planck's constant $(\mathrm{h})=6.6 \times 10^{-34} \mathrm{JS}$ As we know that- $\text { Energy }=\frac{\mathrm{hc}}{\lambda}$ And frequency $(v)=\frac{\mathrm{c}}{\lambda}$ Then, Energy $=\mathrm{h} v$ $=6.6 \times 10^{-34} \mathrm{JS} \times 1.3 \times 10^{15} \mathrm{~s}^{-1}$ $=8.58 \times 10^{-19} \mathrm{~J}$
JEE Main-25.06.2022
Structure of Atom
238684
A particle of mass $6.6 \times 10^{-31} \mathrm{~kg}$ is moving with a velocity of $1 \times 10^7 \mathrm{~ms}^{-1}$. The de Broglie wavelength (in $\AA$ ) associated with the particle, is $\left(\mathrm{h}=6.6 \times 10^{-34} \mathrm{Js}\right)$