142250
The surface of the metal is illuminated with the light of $400 \mathrm{~nm}$. The kinetic energy of the ejected photoelectrons was found to be $1.68 \mathrm{eV}$, the work function of metal is
142252
The diagram shows the energy levels for an electron in a certain atom. Which transition shown represents the emission of a photon with the most energy?
1 III
2 IV
3 I
4 II
Explanation:
A According to the question- $\mathrm{E}=\operatorname{Rhc}\left(\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right)$ Where, $\mathrm{R}$ is the Rydberg's constant $\mathrm{E}_{4 \rightarrow 3} =\operatorname{Rhc}\left[\frac{1}{3^{2}}-\frac{1}{4^{2}}\right]$ $=\operatorname{Rhc}\left[\frac{7}{9 \times 16}\right] \approx 0.05 \mathrm{Rhc}$ $\mathrm{E}_{4 \rightarrow 2} =\operatorname{Rhc}\left[\frac{1}{2^{2}}-\frac{1}{4^{2}}\right]$ $=\operatorname{Rhc} \times\left[\frac{3}{16}\right] \approx 0.2 \mathrm{Rhc}$ $\mathrm{E}_{2 \rightarrow 1} =\operatorname{Rhc}\left[\frac{1}{(1)^{2}}-\frac{1}{(2)^{2}}\right]$ $=\operatorname{Rhc}\left[\frac{3}{4}\right]=0.75 \mathrm{RhC}$ $\mathrm{E}_{1 \rightarrow 3} =\operatorname{Rhc}\left[\frac{1}{(3)^{2}}-\frac{1}{(1)^{2}}\right]$ $=\frac{-8}{9} \mathrm{Rhc} \approx-0.9 \mathrm{Rhc}$ Hence, (III) transition given most energy.
UPSEE - 2007
Dual nature of radiation and Matter
142253
Monochromatic light of frequency $6.0 \times 10^{14}$ Hz. is produced by a laser. The power emitted is $2 \times 10^{-3} \mathrm{~W}$. The number of photons emitted on the average, by the source per second is
1 $5 \times 10^{15}$
2 $5 \times 10^{16}$
3 $5 \times 10^{17}$
4 $5 \times 10^{14}$
Explanation:
A Given that, $\mathrm{P}=2 \times 10^{-3} \mathrm{~W}$ $v=6 \times 10^{14} \mathrm{~Hz}, \mathrm{~h}=6.6 \times 10^{-34} \mathrm{Js}$ The number of photons emitted per second $\mathrm{N}=\frac{\mathrm{P}}{\mathrm{E}}=\frac{2 \times 10^{-3}}{6.63 \times 10^{-34} \times 6 \times 10^{14}} \quad[\because \mathrm{E}=\mathrm{h} v]$ $\mathrm{N}=0.05 \times 10^{17}$ $\mathrm{~N}=5 \times 10^{15}$
UPSEE - 2007
Dual nature of radiation and Matter
142254
Light of energy $2.0 \mathrm{eV}$ falls on a metal of work function $1.4 \mathrm{eV}$. The stopping potential is:
1 $0.6 \mathrm{~V}$
2 $2.0 \mathrm{~V}$
3 $3.4 \mathrm{~V}$
4 $1.4 \mathrm{~V}$
Explanation:
A Given that, Work function $\phi=1.4 \mathrm{eV}$ $\mathrm{E}=\mathrm{hv}=2 \mathrm{eV}$ Since, we know that, $\mathrm{E}_{\mathrm{K}}=\mathrm{h} v-\phi$ $\mathrm{E}_{\mathrm{K}}=2-1.4=0.6 \mathrm{eV}$ Stopping potential, $V_{S}=\frac{E_{K}}{e}$ $=\frac{0.6 \mathrm{eV}}{\mathrm{e}}$ $\mathrm{V}_{\mathrm{S}}=0.6 \mathrm{~V}$
142250
The surface of the metal is illuminated with the light of $400 \mathrm{~nm}$. The kinetic energy of the ejected photoelectrons was found to be $1.68 \mathrm{eV}$, the work function of metal is
142252
The diagram shows the energy levels for an electron in a certain atom. Which transition shown represents the emission of a photon with the most energy?
1 III
2 IV
3 I
4 II
Explanation:
A According to the question- $\mathrm{E}=\operatorname{Rhc}\left(\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right)$ Where, $\mathrm{R}$ is the Rydberg's constant $\mathrm{E}_{4 \rightarrow 3} =\operatorname{Rhc}\left[\frac{1}{3^{2}}-\frac{1}{4^{2}}\right]$ $=\operatorname{Rhc}\left[\frac{7}{9 \times 16}\right] \approx 0.05 \mathrm{Rhc}$ $\mathrm{E}_{4 \rightarrow 2} =\operatorname{Rhc}\left[\frac{1}{2^{2}}-\frac{1}{4^{2}}\right]$ $=\operatorname{Rhc} \times\left[\frac{3}{16}\right] \approx 0.2 \mathrm{Rhc}$ $\mathrm{E}_{2 \rightarrow 1} =\operatorname{Rhc}\left[\frac{1}{(1)^{2}}-\frac{1}{(2)^{2}}\right]$ $=\operatorname{Rhc}\left[\frac{3}{4}\right]=0.75 \mathrm{RhC}$ $\mathrm{E}_{1 \rightarrow 3} =\operatorname{Rhc}\left[\frac{1}{(3)^{2}}-\frac{1}{(1)^{2}}\right]$ $=\frac{-8}{9} \mathrm{Rhc} \approx-0.9 \mathrm{Rhc}$ Hence, (III) transition given most energy.
UPSEE - 2007
Dual nature of radiation and Matter
142253
Monochromatic light of frequency $6.0 \times 10^{14}$ Hz. is produced by a laser. The power emitted is $2 \times 10^{-3} \mathrm{~W}$. The number of photons emitted on the average, by the source per second is
1 $5 \times 10^{15}$
2 $5 \times 10^{16}$
3 $5 \times 10^{17}$
4 $5 \times 10^{14}$
Explanation:
A Given that, $\mathrm{P}=2 \times 10^{-3} \mathrm{~W}$ $v=6 \times 10^{14} \mathrm{~Hz}, \mathrm{~h}=6.6 \times 10^{-34} \mathrm{Js}$ The number of photons emitted per second $\mathrm{N}=\frac{\mathrm{P}}{\mathrm{E}}=\frac{2 \times 10^{-3}}{6.63 \times 10^{-34} \times 6 \times 10^{14}} \quad[\because \mathrm{E}=\mathrm{h} v]$ $\mathrm{N}=0.05 \times 10^{17}$ $\mathrm{~N}=5 \times 10^{15}$
UPSEE - 2007
Dual nature of radiation and Matter
142254
Light of energy $2.0 \mathrm{eV}$ falls on a metal of work function $1.4 \mathrm{eV}$. The stopping potential is:
1 $0.6 \mathrm{~V}$
2 $2.0 \mathrm{~V}$
3 $3.4 \mathrm{~V}$
4 $1.4 \mathrm{~V}$
Explanation:
A Given that, Work function $\phi=1.4 \mathrm{eV}$ $\mathrm{E}=\mathrm{hv}=2 \mathrm{eV}$ Since, we know that, $\mathrm{E}_{\mathrm{K}}=\mathrm{h} v-\phi$ $\mathrm{E}_{\mathrm{K}}=2-1.4=0.6 \mathrm{eV}$ Stopping potential, $V_{S}=\frac{E_{K}}{e}$ $=\frac{0.6 \mathrm{eV}}{\mathrm{e}}$ $\mathrm{V}_{\mathrm{S}}=0.6 \mathrm{~V}$
142250
The surface of the metal is illuminated with the light of $400 \mathrm{~nm}$. The kinetic energy of the ejected photoelectrons was found to be $1.68 \mathrm{eV}$, the work function of metal is
142252
The diagram shows the energy levels for an electron in a certain atom. Which transition shown represents the emission of a photon with the most energy?
1 III
2 IV
3 I
4 II
Explanation:
A According to the question- $\mathrm{E}=\operatorname{Rhc}\left(\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right)$ Where, $\mathrm{R}$ is the Rydberg's constant $\mathrm{E}_{4 \rightarrow 3} =\operatorname{Rhc}\left[\frac{1}{3^{2}}-\frac{1}{4^{2}}\right]$ $=\operatorname{Rhc}\left[\frac{7}{9 \times 16}\right] \approx 0.05 \mathrm{Rhc}$ $\mathrm{E}_{4 \rightarrow 2} =\operatorname{Rhc}\left[\frac{1}{2^{2}}-\frac{1}{4^{2}}\right]$ $=\operatorname{Rhc} \times\left[\frac{3}{16}\right] \approx 0.2 \mathrm{Rhc}$ $\mathrm{E}_{2 \rightarrow 1} =\operatorname{Rhc}\left[\frac{1}{(1)^{2}}-\frac{1}{(2)^{2}}\right]$ $=\operatorname{Rhc}\left[\frac{3}{4}\right]=0.75 \mathrm{RhC}$ $\mathrm{E}_{1 \rightarrow 3} =\operatorname{Rhc}\left[\frac{1}{(3)^{2}}-\frac{1}{(1)^{2}}\right]$ $=\frac{-8}{9} \mathrm{Rhc} \approx-0.9 \mathrm{Rhc}$ Hence, (III) transition given most energy.
UPSEE - 2007
Dual nature of radiation and Matter
142253
Monochromatic light of frequency $6.0 \times 10^{14}$ Hz. is produced by a laser. The power emitted is $2 \times 10^{-3} \mathrm{~W}$. The number of photons emitted on the average, by the source per second is
1 $5 \times 10^{15}$
2 $5 \times 10^{16}$
3 $5 \times 10^{17}$
4 $5 \times 10^{14}$
Explanation:
A Given that, $\mathrm{P}=2 \times 10^{-3} \mathrm{~W}$ $v=6 \times 10^{14} \mathrm{~Hz}, \mathrm{~h}=6.6 \times 10^{-34} \mathrm{Js}$ The number of photons emitted per second $\mathrm{N}=\frac{\mathrm{P}}{\mathrm{E}}=\frac{2 \times 10^{-3}}{6.63 \times 10^{-34} \times 6 \times 10^{14}} \quad[\because \mathrm{E}=\mathrm{h} v]$ $\mathrm{N}=0.05 \times 10^{17}$ $\mathrm{~N}=5 \times 10^{15}$
UPSEE - 2007
Dual nature of radiation and Matter
142254
Light of energy $2.0 \mathrm{eV}$ falls on a metal of work function $1.4 \mathrm{eV}$. The stopping potential is:
1 $0.6 \mathrm{~V}$
2 $2.0 \mathrm{~V}$
3 $3.4 \mathrm{~V}$
4 $1.4 \mathrm{~V}$
Explanation:
A Given that, Work function $\phi=1.4 \mathrm{eV}$ $\mathrm{E}=\mathrm{hv}=2 \mathrm{eV}$ Since, we know that, $\mathrm{E}_{\mathrm{K}}=\mathrm{h} v-\phi$ $\mathrm{E}_{\mathrm{K}}=2-1.4=0.6 \mathrm{eV}$ Stopping potential, $V_{S}=\frac{E_{K}}{e}$ $=\frac{0.6 \mathrm{eV}}{\mathrm{e}}$ $\mathrm{V}_{\mathrm{S}}=0.6 \mathrm{~V}$
142250
The surface of the metal is illuminated with the light of $400 \mathrm{~nm}$. The kinetic energy of the ejected photoelectrons was found to be $1.68 \mathrm{eV}$, the work function of metal is
142252
The diagram shows the energy levels for an electron in a certain atom. Which transition shown represents the emission of a photon with the most energy?
1 III
2 IV
3 I
4 II
Explanation:
A According to the question- $\mathrm{E}=\operatorname{Rhc}\left(\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right)$ Where, $\mathrm{R}$ is the Rydberg's constant $\mathrm{E}_{4 \rightarrow 3} =\operatorname{Rhc}\left[\frac{1}{3^{2}}-\frac{1}{4^{2}}\right]$ $=\operatorname{Rhc}\left[\frac{7}{9 \times 16}\right] \approx 0.05 \mathrm{Rhc}$ $\mathrm{E}_{4 \rightarrow 2} =\operatorname{Rhc}\left[\frac{1}{2^{2}}-\frac{1}{4^{2}}\right]$ $=\operatorname{Rhc} \times\left[\frac{3}{16}\right] \approx 0.2 \mathrm{Rhc}$ $\mathrm{E}_{2 \rightarrow 1} =\operatorname{Rhc}\left[\frac{1}{(1)^{2}}-\frac{1}{(2)^{2}}\right]$ $=\operatorname{Rhc}\left[\frac{3}{4}\right]=0.75 \mathrm{RhC}$ $\mathrm{E}_{1 \rightarrow 3} =\operatorname{Rhc}\left[\frac{1}{(3)^{2}}-\frac{1}{(1)^{2}}\right]$ $=\frac{-8}{9} \mathrm{Rhc} \approx-0.9 \mathrm{Rhc}$ Hence, (III) transition given most energy.
UPSEE - 2007
Dual nature of radiation and Matter
142253
Monochromatic light of frequency $6.0 \times 10^{14}$ Hz. is produced by a laser. The power emitted is $2 \times 10^{-3} \mathrm{~W}$. The number of photons emitted on the average, by the source per second is
1 $5 \times 10^{15}$
2 $5 \times 10^{16}$
3 $5 \times 10^{17}$
4 $5 \times 10^{14}$
Explanation:
A Given that, $\mathrm{P}=2 \times 10^{-3} \mathrm{~W}$ $v=6 \times 10^{14} \mathrm{~Hz}, \mathrm{~h}=6.6 \times 10^{-34} \mathrm{Js}$ The number of photons emitted per second $\mathrm{N}=\frac{\mathrm{P}}{\mathrm{E}}=\frac{2 \times 10^{-3}}{6.63 \times 10^{-34} \times 6 \times 10^{14}} \quad[\because \mathrm{E}=\mathrm{h} v]$ $\mathrm{N}=0.05 \times 10^{17}$ $\mathrm{~N}=5 \times 10^{15}$
UPSEE - 2007
Dual nature of radiation and Matter
142254
Light of energy $2.0 \mathrm{eV}$ falls on a metal of work function $1.4 \mathrm{eV}$. The stopping potential is:
1 $0.6 \mathrm{~V}$
2 $2.0 \mathrm{~V}$
3 $3.4 \mathrm{~V}$
4 $1.4 \mathrm{~V}$
Explanation:
A Given that, Work function $\phi=1.4 \mathrm{eV}$ $\mathrm{E}=\mathrm{hv}=2 \mathrm{eV}$ Since, we know that, $\mathrm{E}_{\mathrm{K}}=\mathrm{h} v-\phi$ $\mathrm{E}_{\mathrm{K}}=2-1.4=0.6 \mathrm{eV}$ Stopping potential, $V_{S}=\frac{E_{K}}{e}$ $=\frac{0.6 \mathrm{eV}}{\mathrm{e}}$ $\mathrm{V}_{\mathrm{S}}=0.6 \mathrm{~V}$
