357799
The photo electrons emitted from the surface of sodium metal are
1 Of speeds from 0 to a certain maximum
2 Of same de Broglie wavelength
3 Of same kinetic energy
4 Of same frequency
Explanation:
Conceptual Question
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357800
Stopping potential depends on
1 Frequency of incident light
2 Intensity of incident light
3 Number of emitted electrons
4 Number of incident photons
Explanation:
Conceptual Question
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357801
The function of photoelectric cell is
1 To convert electrical energy into light energy
2 To convert light energy into electrical energy
3 To convert mechanical energy into electrical energy
4 To convert \(DC\) into \(AC\)
Explanation:
Conceptual Question
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357802
Wavelength of a 1 \(KeV\) photon is \({1.24 \times 10^{-9} {~m}}\). What is the frequency of 1 \(KeV\) photon?
1 \({2.4 \times 10^{20} {~Hz}}\)
2 \({1.24 \times 10^{15} {~Hz}}\)
3 \({1.24 \times 10^{18} {~Hz}}\)
4 \({2.4 \times 10^{23} {~Hz}}\)
Explanation:
Energy of the photon is given by \({E=h v}\) Here \({h=}\) Planck's constant \({v=}\) frequency of photon \({{v}=\dfrac{E}{h}=\dfrac{10^{6} {eV}}{6.6 \times 10^{-34} {Js}}}\) \({v=\dfrac{10^{6} \times 1.6 \times 10^{-19}}{6.6 \times 10^{-34}} {~Hz}}\) \(v = 0.2424 \times {10^{21}}\;Hz = 2.4 \times {10^{20}}\;Hz\). So correct option is (1)
357799
The photo electrons emitted from the surface of sodium metal are
1 Of speeds from 0 to a certain maximum
2 Of same de Broglie wavelength
3 Of same kinetic energy
4 Of same frequency
Explanation:
Conceptual Question
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357800
Stopping potential depends on
1 Frequency of incident light
2 Intensity of incident light
3 Number of emitted electrons
4 Number of incident photons
Explanation:
Conceptual Question
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357801
The function of photoelectric cell is
1 To convert electrical energy into light energy
2 To convert light energy into electrical energy
3 To convert mechanical energy into electrical energy
4 To convert \(DC\) into \(AC\)
Explanation:
Conceptual Question
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357802
Wavelength of a 1 \(KeV\) photon is \({1.24 \times 10^{-9} {~m}}\). What is the frequency of 1 \(KeV\) photon?
1 \({2.4 \times 10^{20} {~Hz}}\)
2 \({1.24 \times 10^{15} {~Hz}}\)
3 \({1.24 \times 10^{18} {~Hz}}\)
4 \({2.4 \times 10^{23} {~Hz}}\)
Explanation:
Energy of the photon is given by \({E=h v}\) Here \({h=}\) Planck's constant \({v=}\) frequency of photon \({{v}=\dfrac{E}{h}=\dfrac{10^{6} {eV}}{6.6 \times 10^{-34} {Js}}}\) \({v=\dfrac{10^{6} \times 1.6 \times 10^{-19}}{6.6 \times 10^{-34}} {~Hz}}\) \(v = 0.2424 \times {10^{21}}\;Hz = 2.4 \times {10^{20}}\;Hz\). So correct option is (1)
357799
The photo electrons emitted from the surface of sodium metal are
1 Of speeds from 0 to a certain maximum
2 Of same de Broglie wavelength
3 Of same kinetic energy
4 Of same frequency
Explanation:
Conceptual Question
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357800
Stopping potential depends on
1 Frequency of incident light
2 Intensity of incident light
3 Number of emitted electrons
4 Number of incident photons
Explanation:
Conceptual Question
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357801
The function of photoelectric cell is
1 To convert electrical energy into light energy
2 To convert light energy into electrical energy
3 To convert mechanical energy into electrical energy
4 To convert \(DC\) into \(AC\)
Explanation:
Conceptual Question
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357802
Wavelength of a 1 \(KeV\) photon is \({1.24 \times 10^{-9} {~m}}\). What is the frequency of 1 \(KeV\) photon?
1 \({2.4 \times 10^{20} {~Hz}}\)
2 \({1.24 \times 10^{15} {~Hz}}\)
3 \({1.24 \times 10^{18} {~Hz}}\)
4 \({2.4 \times 10^{23} {~Hz}}\)
Explanation:
Energy of the photon is given by \({E=h v}\) Here \({h=}\) Planck's constant \({v=}\) frequency of photon \({{v}=\dfrac{E}{h}=\dfrac{10^{6} {eV}}{6.6 \times 10^{-34} {Js}}}\) \({v=\dfrac{10^{6} \times 1.6 \times 10^{-19}}{6.6 \times 10^{-34}} {~Hz}}\) \(v = 0.2424 \times {10^{21}}\;Hz = 2.4 \times {10^{20}}\;Hz\). So correct option is (1)
357799
The photo electrons emitted from the surface of sodium metal are
1 Of speeds from 0 to a certain maximum
2 Of same de Broglie wavelength
3 Of same kinetic energy
4 Of same frequency
Explanation:
Conceptual Question
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357800
Stopping potential depends on
1 Frequency of incident light
2 Intensity of incident light
3 Number of emitted electrons
4 Number of incident photons
Explanation:
Conceptual Question
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357801
The function of photoelectric cell is
1 To convert electrical energy into light energy
2 To convert light energy into electrical energy
3 To convert mechanical energy into electrical energy
4 To convert \(DC\) into \(AC\)
Explanation:
Conceptual Question
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357802
Wavelength of a 1 \(KeV\) photon is \({1.24 \times 10^{-9} {~m}}\). What is the frequency of 1 \(KeV\) photon?
1 \({2.4 \times 10^{20} {~Hz}}\)
2 \({1.24 \times 10^{15} {~Hz}}\)
3 \({1.24 \times 10^{18} {~Hz}}\)
4 \({2.4 \times 10^{23} {~Hz}}\)
Explanation:
Energy of the photon is given by \({E=h v}\) Here \({h=}\) Planck's constant \({v=}\) frequency of photon \({{v}=\dfrac{E}{h}=\dfrac{10^{6} {eV}}{6.6 \times 10^{-34} {Js}}}\) \({v=\dfrac{10^{6} \times 1.6 \times 10^{-19}}{6.6 \times 10^{-34}} {~Hz}}\) \(v = 0.2424 \times {10^{21}}\;Hz = 2.4 \times {10^{20}}\;Hz\). So correct option is (1)
357799
The photo electrons emitted from the surface of sodium metal are
1 Of speeds from 0 to a certain maximum
2 Of same de Broglie wavelength
3 Of same kinetic energy
4 Of same frequency
Explanation:
Conceptual Question
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357800
Stopping potential depends on
1 Frequency of incident light
2 Intensity of incident light
3 Number of emitted electrons
4 Number of incident photons
Explanation:
Conceptual Question
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357801
The function of photoelectric cell is
1 To convert electrical energy into light energy
2 To convert light energy into electrical energy
3 To convert mechanical energy into electrical energy
4 To convert \(DC\) into \(AC\)
Explanation:
Conceptual Question
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357802
Wavelength of a 1 \(KeV\) photon is \({1.24 \times 10^{-9} {~m}}\). What is the frequency of 1 \(KeV\) photon?
1 \({2.4 \times 10^{20} {~Hz}}\)
2 \({1.24 \times 10^{15} {~Hz}}\)
3 \({1.24 \times 10^{18} {~Hz}}\)
4 \({2.4 \times 10^{23} {~Hz}}\)
Explanation:
Energy of the photon is given by \({E=h v}\) Here \({h=}\) Planck's constant \({v=}\) frequency of photon \({{v}=\dfrac{E}{h}=\dfrac{10^{6} {eV}}{6.6 \times 10^{-34} {Js}}}\) \({v=\dfrac{10^{6} \times 1.6 \times 10^{-19}}{6.6 \times 10^{-34}} {~Hz}}\) \(v = 0.2424 \times {10^{21}}\;Hz = 2.4 \times {10^{20}}\;Hz\). So correct option is (1)
