142245
In a photoelectric experiment, with light of wavelength $\lambda$, the fastest electron has speed $v$. If the exciting wavelength is changed to $3 \lambda / 4$, the speed of the fastest emitted electron will become
1 $v \sqrt{\frac{3}{4}}$
2 $v \sqrt{\frac{4}{3}}$
3 less than $v \sqrt{\frac{4}{3}}$
4 greater than $v \sqrt{\frac{4}{3}}$
Explanation:
D Given that, $\lambda=\frac{3 \lambda}{4}$ We know that, $(\mathrm{K} . \mathrm{E})_{\max }=\frac{\mathrm{hc}}{\lambda}-\phi_{0}$ $\frac{1}{2} \mathrm{mv}^{2}=\frac{\mathrm{hc}}{\lambda}-\phi_{0}$ $\frac{1}{2} \mathrm{~m}\left(\mathrm{v}^{\prime}\right)^{2}=\frac{\mathrm{hc}}{3 \lambda / 4}-\phi_{0}$ $\frac{1}{2} \mathrm{~m}\left(\mathrm{v}^{\prime}\right)^{2}=\frac{4 \mathrm{hc}}{3 \lambda}-\phi_{0}$ Dividing equation (ii) by equation (i), we get- $\therefore \quad \frac{\frac{1}{2} \mathrm{mv}^{\prime 2}}{\frac{1}{2} \mathrm{~m} \mathrm{v}^{2}}=\frac{\frac{4 \mathrm{hc}}{3 \lambda}-\phi_{0}+\frac{4}{3} \phi_{0}-\frac{4}{3} \phi_{0}}{\frac{\mathrm{hc}}{\lambda}-\phi_{0}}$ $\frac{\mathrm{v}^{\prime 2}}{\mathrm{v}^{2}}=\frac{\frac{4 \mathrm{hc}}{3 \lambda}-\frac{4}{3} \phi_{0}+\frac{4}{3} \phi_{0}-\phi_{0}}{\frac{\mathrm{hc}}{\lambda}-\phi_{0}}$ $\frac{\mathrm{v}^{\prime 2}}{\mathrm{v}^{2}}=\frac{\frac{4}{3}\left(\frac{\mathrm{hc}}{\lambda}-\phi_{0}\right)+\frac{\phi_{0}}{3}}{\frac{\mathrm{hc}}{\lambda}-\phi_{0}}$ $\frac{\mathrm{v}^{\prime^{2}}}{\mathrm{v}^{2}}=\frac{4}{3}+\frac{\phi_{0} / 3}{\frac{\mathrm{hc}}{\lambda}-\phi}>\frac{4}{3}$ $\frac{\mathrm{v}^{\prime 2}}{\mathrm{v}^{2}}>\frac{4}{3}$ $\mathrm{v}^{\prime}>\sqrt{\frac{4}{3}} \mathrm{v}$ $\mathrm{v}^{\prime}>\sqrt{\frac{4}{3}} \mathrm{v}$
BITSAT-2009
Dual nature of radiation and Matter
142246
The threshold wavelength of the tungsten is $2300 \AA$. If ultraviolet light of wavelength 1800 $\AA$ is incident on it, then the maximum kinetic energy of photoelectrons would be about -
142248
A light having wavelength $300 \mathrm{~nm}$ fall on a metal surface. The work function of metal is $2.54 \mathrm{eV}$, what is stopping potential?
1 $2.3 \mathrm{~V}$
2 $2.59 \mathrm{~V}$
3 $1.59 \mathrm{~V}$
4 $1.29 \mathrm{~V}$
Explanation:
C Given that, $\lambda=300 \mathrm{~nm}$ $\phi=2.54 \mathrm{eV}$ We know that, From Einstein's photoelectric equation- $\frac{\mathrm{hc}}{\lambda}-\phi=\mathrm{eV}_{\mathrm{o}}$ $\frac{\left(6.62 \times 10^{-34} \times 3 \times 10^{8}\right)}{1.6 \times 10^{-19} \times 3 \times 10^{-7}}-2.54=\mathrm{eV}_{\mathrm{o}}$ $4.137-2.54=\mathrm{eV}_{\mathrm{o}}$ $\mathrm{V}_{0}=1.6 \mathrm{~V} \sqcup 1.59 \mathrm{~V}$
BITSAT-2012
Dual nature of radiation and Matter
142249
The postulate on which the photoelectric equation is derived is
1 light is emitted only when electrons jump between orbits.
2 light is absorbed in quanta of energy $E=h v$
3 electrons are restricted to orbits of angular momentum $n \frac{h}{2 \pi}$ where $n$ is an integer.
4 electrons are associated with wave of wavelength $\lambda=\frac{\mathrm{h}}{\mathrm{p}}$ where $\mathrm{p}$ is momentum.
Explanation:
B Photo electric effect, $\mathrm{E}=\mathrm{h} v=\frac{\mathrm{hc}}{\lambda}$ Absorbed metal surface $\mathrm{K}_{\max }=\mathrm{h} v-\phi$ Where, $\phi$ work function $\mathrm{h}$ is Planck's constant. This can be understand by in terms of particle nature of light.
NEET Test Series from KOTA - 10 Papers In MS WORD
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Dual nature of radiation and Matter
142245
In a photoelectric experiment, with light of wavelength $\lambda$, the fastest electron has speed $v$. If the exciting wavelength is changed to $3 \lambda / 4$, the speed of the fastest emitted electron will become
1 $v \sqrt{\frac{3}{4}}$
2 $v \sqrt{\frac{4}{3}}$
3 less than $v \sqrt{\frac{4}{3}}$
4 greater than $v \sqrt{\frac{4}{3}}$
Explanation:
D Given that, $\lambda=\frac{3 \lambda}{4}$ We know that, $(\mathrm{K} . \mathrm{E})_{\max }=\frac{\mathrm{hc}}{\lambda}-\phi_{0}$ $\frac{1}{2} \mathrm{mv}^{2}=\frac{\mathrm{hc}}{\lambda}-\phi_{0}$ $\frac{1}{2} \mathrm{~m}\left(\mathrm{v}^{\prime}\right)^{2}=\frac{\mathrm{hc}}{3 \lambda / 4}-\phi_{0}$ $\frac{1}{2} \mathrm{~m}\left(\mathrm{v}^{\prime}\right)^{2}=\frac{4 \mathrm{hc}}{3 \lambda}-\phi_{0}$ Dividing equation (ii) by equation (i), we get- $\therefore \quad \frac{\frac{1}{2} \mathrm{mv}^{\prime 2}}{\frac{1}{2} \mathrm{~m} \mathrm{v}^{2}}=\frac{\frac{4 \mathrm{hc}}{3 \lambda}-\phi_{0}+\frac{4}{3} \phi_{0}-\frac{4}{3} \phi_{0}}{\frac{\mathrm{hc}}{\lambda}-\phi_{0}}$ $\frac{\mathrm{v}^{\prime 2}}{\mathrm{v}^{2}}=\frac{\frac{4 \mathrm{hc}}{3 \lambda}-\frac{4}{3} \phi_{0}+\frac{4}{3} \phi_{0}-\phi_{0}}{\frac{\mathrm{hc}}{\lambda}-\phi_{0}}$ $\frac{\mathrm{v}^{\prime 2}}{\mathrm{v}^{2}}=\frac{\frac{4}{3}\left(\frac{\mathrm{hc}}{\lambda}-\phi_{0}\right)+\frac{\phi_{0}}{3}}{\frac{\mathrm{hc}}{\lambda}-\phi_{0}}$ $\frac{\mathrm{v}^{\prime^{2}}}{\mathrm{v}^{2}}=\frac{4}{3}+\frac{\phi_{0} / 3}{\frac{\mathrm{hc}}{\lambda}-\phi}>\frac{4}{3}$ $\frac{\mathrm{v}^{\prime 2}}{\mathrm{v}^{2}}>\frac{4}{3}$ $\mathrm{v}^{\prime}>\sqrt{\frac{4}{3}} \mathrm{v}$ $\mathrm{v}^{\prime}>\sqrt{\frac{4}{3}} \mathrm{v}$
BITSAT-2009
Dual nature of radiation and Matter
142246
The threshold wavelength of the tungsten is $2300 \AA$. If ultraviolet light of wavelength 1800 $\AA$ is incident on it, then the maximum kinetic energy of photoelectrons would be about -
142248
A light having wavelength $300 \mathrm{~nm}$ fall on a metal surface. The work function of metal is $2.54 \mathrm{eV}$, what is stopping potential?
1 $2.3 \mathrm{~V}$
2 $2.59 \mathrm{~V}$
3 $1.59 \mathrm{~V}$
4 $1.29 \mathrm{~V}$
Explanation:
C Given that, $\lambda=300 \mathrm{~nm}$ $\phi=2.54 \mathrm{eV}$ We know that, From Einstein's photoelectric equation- $\frac{\mathrm{hc}}{\lambda}-\phi=\mathrm{eV}_{\mathrm{o}}$ $\frac{\left(6.62 \times 10^{-34} \times 3 \times 10^{8}\right)}{1.6 \times 10^{-19} \times 3 \times 10^{-7}}-2.54=\mathrm{eV}_{\mathrm{o}}$ $4.137-2.54=\mathrm{eV}_{\mathrm{o}}$ $\mathrm{V}_{0}=1.6 \mathrm{~V} \sqcup 1.59 \mathrm{~V}$
BITSAT-2012
Dual nature of radiation and Matter
142249
The postulate on which the photoelectric equation is derived is
1 light is emitted only when electrons jump between orbits.
2 light is absorbed in quanta of energy $E=h v$
3 electrons are restricted to orbits of angular momentum $n \frac{h}{2 \pi}$ where $n$ is an integer.
4 electrons are associated with wave of wavelength $\lambda=\frac{\mathrm{h}}{\mathrm{p}}$ where $\mathrm{p}$ is momentum.
Explanation:
B Photo electric effect, $\mathrm{E}=\mathrm{h} v=\frac{\mathrm{hc}}{\lambda}$ Absorbed metal surface $\mathrm{K}_{\max }=\mathrm{h} v-\phi$ Where, $\phi$ work function $\mathrm{h}$ is Planck's constant. This can be understand by in terms of particle nature of light.
142245
In a photoelectric experiment, with light of wavelength $\lambda$, the fastest electron has speed $v$. If the exciting wavelength is changed to $3 \lambda / 4$, the speed of the fastest emitted electron will become
1 $v \sqrt{\frac{3}{4}}$
2 $v \sqrt{\frac{4}{3}}$
3 less than $v \sqrt{\frac{4}{3}}$
4 greater than $v \sqrt{\frac{4}{3}}$
Explanation:
D Given that, $\lambda=\frac{3 \lambda}{4}$ We know that, $(\mathrm{K} . \mathrm{E})_{\max }=\frac{\mathrm{hc}}{\lambda}-\phi_{0}$ $\frac{1}{2} \mathrm{mv}^{2}=\frac{\mathrm{hc}}{\lambda}-\phi_{0}$ $\frac{1}{2} \mathrm{~m}\left(\mathrm{v}^{\prime}\right)^{2}=\frac{\mathrm{hc}}{3 \lambda / 4}-\phi_{0}$ $\frac{1}{2} \mathrm{~m}\left(\mathrm{v}^{\prime}\right)^{2}=\frac{4 \mathrm{hc}}{3 \lambda}-\phi_{0}$ Dividing equation (ii) by equation (i), we get- $\therefore \quad \frac{\frac{1}{2} \mathrm{mv}^{\prime 2}}{\frac{1}{2} \mathrm{~m} \mathrm{v}^{2}}=\frac{\frac{4 \mathrm{hc}}{3 \lambda}-\phi_{0}+\frac{4}{3} \phi_{0}-\frac{4}{3} \phi_{0}}{\frac{\mathrm{hc}}{\lambda}-\phi_{0}}$ $\frac{\mathrm{v}^{\prime 2}}{\mathrm{v}^{2}}=\frac{\frac{4 \mathrm{hc}}{3 \lambda}-\frac{4}{3} \phi_{0}+\frac{4}{3} \phi_{0}-\phi_{0}}{\frac{\mathrm{hc}}{\lambda}-\phi_{0}}$ $\frac{\mathrm{v}^{\prime 2}}{\mathrm{v}^{2}}=\frac{\frac{4}{3}\left(\frac{\mathrm{hc}}{\lambda}-\phi_{0}\right)+\frac{\phi_{0}}{3}}{\frac{\mathrm{hc}}{\lambda}-\phi_{0}}$ $\frac{\mathrm{v}^{\prime^{2}}}{\mathrm{v}^{2}}=\frac{4}{3}+\frac{\phi_{0} / 3}{\frac{\mathrm{hc}}{\lambda}-\phi}>\frac{4}{3}$ $\frac{\mathrm{v}^{\prime 2}}{\mathrm{v}^{2}}>\frac{4}{3}$ $\mathrm{v}^{\prime}>\sqrt{\frac{4}{3}} \mathrm{v}$ $\mathrm{v}^{\prime}>\sqrt{\frac{4}{3}} \mathrm{v}$
BITSAT-2009
Dual nature of radiation and Matter
142246
The threshold wavelength of the tungsten is $2300 \AA$. If ultraviolet light of wavelength 1800 $\AA$ is incident on it, then the maximum kinetic energy of photoelectrons would be about -
142248
A light having wavelength $300 \mathrm{~nm}$ fall on a metal surface. The work function of metal is $2.54 \mathrm{eV}$, what is stopping potential?
1 $2.3 \mathrm{~V}$
2 $2.59 \mathrm{~V}$
3 $1.59 \mathrm{~V}$
4 $1.29 \mathrm{~V}$
Explanation:
C Given that, $\lambda=300 \mathrm{~nm}$ $\phi=2.54 \mathrm{eV}$ We know that, From Einstein's photoelectric equation- $\frac{\mathrm{hc}}{\lambda}-\phi=\mathrm{eV}_{\mathrm{o}}$ $\frac{\left(6.62 \times 10^{-34} \times 3 \times 10^{8}\right)}{1.6 \times 10^{-19} \times 3 \times 10^{-7}}-2.54=\mathrm{eV}_{\mathrm{o}}$ $4.137-2.54=\mathrm{eV}_{\mathrm{o}}$ $\mathrm{V}_{0}=1.6 \mathrm{~V} \sqcup 1.59 \mathrm{~V}$
BITSAT-2012
Dual nature of radiation and Matter
142249
The postulate on which the photoelectric equation is derived is
1 light is emitted only when electrons jump between orbits.
2 light is absorbed in quanta of energy $E=h v$
3 electrons are restricted to orbits of angular momentum $n \frac{h}{2 \pi}$ where $n$ is an integer.
4 electrons are associated with wave of wavelength $\lambda=\frac{\mathrm{h}}{\mathrm{p}}$ where $\mathrm{p}$ is momentum.
Explanation:
B Photo electric effect, $\mathrm{E}=\mathrm{h} v=\frac{\mathrm{hc}}{\lambda}$ Absorbed metal surface $\mathrm{K}_{\max }=\mathrm{h} v-\phi$ Where, $\phi$ work function $\mathrm{h}$ is Planck's constant. This can be understand by in terms of particle nature of light.
142245
In a photoelectric experiment, with light of wavelength $\lambda$, the fastest electron has speed $v$. If the exciting wavelength is changed to $3 \lambda / 4$, the speed of the fastest emitted electron will become
1 $v \sqrt{\frac{3}{4}}$
2 $v \sqrt{\frac{4}{3}}$
3 less than $v \sqrt{\frac{4}{3}}$
4 greater than $v \sqrt{\frac{4}{3}}$
Explanation:
D Given that, $\lambda=\frac{3 \lambda}{4}$ We know that, $(\mathrm{K} . \mathrm{E})_{\max }=\frac{\mathrm{hc}}{\lambda}-\phi_{0}$ $\frac{1}{2} \mathrm{mv}^{2}=\frac{\mathrm{hc}}{\lambda}-\phi_{0}$ $\frac{1}{2} \mathrm{~m}\left(\mathrm{v}^{\prime}\right)^{2}=\frac{\mathrm{hc}}{3 \lambda / 4}-\phi_{0}$ $\frac{1}{2} \mathrm{~m}\left(\mathrm{v}^{\prime}\right)^{2}=\frac{4 \mathrm{hc}}{3 \lambda}-\phi_{0}$ Dividing equation (ii) by equation (i), we get- $\therefore \quad \frac{\frac{1}{2} \mathrm{mv}^{\prime 2}}{\frac{1}{2} \mathrm{~m} \mathrm{v}^{2}}=\frac{\frac{4 \mathrm{hc}}{3 \lambda}-\phi_{0}+\frac{4}{3} \phi_{0}-\frac{4}{3} \phi_{0}}{\frac{\mathrm{hc}}{\lambda}-\phi_{0}}$ $\frac{\mathrm{v}^{\prime 2}}{\mathrm{v}^{2}}=\frac{\frac{4 \mathrm{hc}}{3 \lambda}-\frac{4}{3} \phi_{0}+\frac{4}{3} \phi_{0}-\phi_{0}}{\frac{\mathrm{hc}}{\lambda}-\phi_{0}}$ $\frac{\mathrm{v}^{\prime 2}}{\mathrm{v}^{2}}=\frac{\frac{4}{3}\left(\frac{\mathrm{hc}}{\lambda}-\phi_{0}\right)+\frac{\phi_{0}}{3}}{\frac{\mathrm{hc}}{\lambda}-\phi_{0}}$ $\frac{\mathrm{v}^{\prime^{2}}}{\mathrm{v}^{2}}=\frac{4}{3}+\frac{\phi_{0} / 3}{\frac{\mathrm{hc}}{\lambda}-\phi}>\frac{4}{3}$ $\frac{\mathrm{v}^{\prime 2}}{\mathrm{v}^{2}}>\frac{4}{3}$ $\mathrm{v}^{\prime}>\sqrt{\frac{4}{3}} \mathrm{v}$ $\mathrm{v}^{\prime}>\sqrt{\frac{4}{3}} \mathrm{v}$
BITSAT-2009
Dual nature of radiation and Matter
142246
The threshold wavelength of the tungsten is $2300 \AA$. If ultraviolet light of wavelength 1800 $\AA$ is incident on it, then the maximum kinetic energy of photoelectrons would be about -
142248
A light having wavelength $300 \mathrm{~nm}$ fall on a metal surface. The work function of metal is $2.54 \mathrm{eV}$, what is stopping potential?
1 $2.3 \mathrm{~V}$
2 $2.59 \mathrm{~V}$
3 $1.59 \mathrm{~V}$
4 $1.29 \mathrm{~V}$
Explanation:
C Given that, $\lambda=300 \mathrm{~nm}$ $\phi=2.54 \mathrm{eV}$ We know that, From Einstein's photoelectric equation- $\frac{\mathrm{hc}}{\lambda}-\phi=\mathrm{eV}_{\mathrm{o}}$ $\frac{\left(6.62 \times 10^{-34} \times 3 \times 10^{8}\right)}{1.6 \times 10^{-19} \times 3 \times 10^{-7}}-2.54=\mathrm{eV}_{\mathrm{o}}$ $4.137-2.54=\mathrm{eV}_{\mathrm{o}}$ $\mathrm{V}_{0}=1.6 \mathrm{~V} \sqcup 1.59 \mathrm{~V}$
BITSAT-2012
Dual nature of radiation and Matter
142249
The postulate on which the photoelectric equation is derived is
1 light is emitted only when electrons jump between orbits.
2 light is absorbed in quanta of energy $E=h v$
3 electrons are restricted to orbits of angular momentum $n \frac{h}{2 \pi}$ where $n$ is an integer.
4 electrons are associated with wave of wavelength $\lambda=\frac{\mathrm{h}}{\mathrm{p}}$ where $\mathrm{p}$ is momentum.
Explanation:
B Photo electric effect, $\mathrm{E}=\mathrm{h} v=\frac{\mathrm{hc}}{\lambda}$ Absorbed metal surface $\mathrm{K}_{\max }=\mathrm{h} v-\phi$ Where, $\phi$ work function $\mathrm{h}$ is Planck's constant. This can be understand by in terms of particle nature of light.
