357751
For the photoelectric effect, the maximum kinetic energy \(E_{k}\) of the emitted photoelectrons is plotted against the frequency \(v\) of the incident photons as shown in the figure. The slope of the curve gives
1 Charge of the electron
2 Work function of the metal
3 Planck's constant
4 Ratio of the Planck's constant to electronic charge
Explanation:
Comparing Einstein's equation \(K_{\max }=h v-\phi\), with \(y=m x+c\), we get slope, \(m=h\)
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357752
A photometal is illuminated by lights of wavelengths \(\lambda_{1}\) and \(\lambda_{2}\) respectively. The maximum kinetic energies of electrons emitted in the two cases are \(E_{1}\) and \(E_{2}\) respectively. The work function of metal is.
357753
The maximum kinetic energy of the photoelectrons depends only on
1 Incident angle
2 Potential
3 Pressure
4 Frequency
Explanation:
using \({E_k} = hv - h{v_0},\) maximum kinetic energy of photoelctrons depends only on frequency.
KCET - 2014
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357754
The correct graph between the maximum energy of a photoelectron and the inverse of wavelength of the incident radiation is given by the curve
1 \(A\)
2 \(B\)
3 \(C\)
4 None of the above
Explanation:
\(K_{\max }=h v-\phi=\dfrac{h c}{\lambda}-\phi\) i.e., graph between \(K_{\max }\) and \(\dfrac{1}{\lambda}\) will be straight line having slope (\(hc\)) and intercept \(\phi\) on -\(KE\) axis. So option (1) is correct.
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357755
Light of energy \({E}\) falls normally on a metal of work function \({E / 3}\). The kinetic energies \({(K)}\) of the photo electrons are
1 \({K=\dfrac{2 E}{3}}\)
2 \({K=\dfrac{E}{3}}\)
3 \({0 \leq K \leq \dfrac{2 E}{3}}\)
4 \({0 \leq K \leq \dfrac{E}{3}}\)
Explanation:
From Einstein's equation for photoelectric emission, maximum kinetic energy of photo electrons is given by \(K{E_{\max }} = E - {\omega _0}\) \( = E - \frac{E}{3} = \frac{{2E}}{3}\) \({\therefore K E}\) lies from 0 to \({\dfrac{2 E}{3}}\). Correct option is (3).
357751
For the photoelectric effect, the maximum kinetic energy \(E_{k}\) of the emitted photoelectrons is plotted against the frequency \(v\) of the incident photons as shown in the figure. The slope of the curve gives
1 Charge of the electron
2 Work function of the metal
3 Planck's constant
4 Ratio of the Planck's constant to electronic charge
Explanation:
Comparing Einstein's equation \(K_{\max }=h v-\phi\), with \(y=m x+c\), we get slope, \(m=h\)
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357752
A photometal is illuminated by lights of wavelengths \(\lambda_{1}\) and \(\lambda_{2}\) respectively. The maximum kinetic energies of electrons emitted in the two cases are \(E_{1}\) and \(E_{2}\) respectively. The work function of metal is.
357753
The maximum kinetic energy of the photoelectrons depends only on
1 Incident angle
2 Potential
3 Pressure
4 Frequency
Explanation:
using \({E_k} = hv - h{v_0},\) maximum kinetic energy of photoelctrons depends only on frequency.
KCET - 2014
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357754
The correct graph between the maximum energy of a photoelectron and the inverse of wavelength of the incident radiation is given by the curve
1 \(A\)
2 \(B\)
3 \(C\)
4 None of the above
Explanation:
\(K_{\max }=h v-\phi=\dfrac{h c}{\lambda}-\phi\) i.e., graph between \(K_{\max }\) and \(\dfrac{1}{\lambda}\) will be straight line having slope (\(hc\)) and intercept \(\phi\) on -\(KE\) axis. So option (1) is correct.
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357755
Light of energy \({E}\) falls normally on a metal of work function \({E / 3}\). The kinetic energies \({(K)}\) of the photo electrons are
1 \({K=\dfrac{2 E}{3}}\)
2 \({K=\dfrac{E}{3}}\)
3 \({0 \leq K \leq \dfrac{2 E}{3}}\)
4 \({0 \leq K \leq \dfrac{E}{3}}\)
Explanation:
From Einstein's equation for photoelectric emission, maximum kinetic energy of photo electrons is given by \(K{E_{\max }} = E - {\omega _0}\) \( = E - \frac{E}{3} = \frac{{2E}}{3}\) \({\therefore K E}\) lies from 0 to \({\dfrac{2 E}{3}}\). Correct option is (3).
NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXII11:DUAL NATURE OF RADIATION AND MATTER
357751
For the photoelectric effect, the maximum kinetic energy \(E_{k}\) of the emitted photoelectrons is plotted against the frequency \(v\) of the incident photons as shown in the figure. The slope of the curve gives
1 Charge of the electron
2 Work function of the metal
3 Planck's constant
4 Ratio of the Planck's constant to electronic charge
Explanation:
Comparing Einstein's equation \(K_{\max }=h v-\phi\), with \(y=m x+c\), we get slope, \(m=h\)
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357752
A photometal is illuminated by lights of wavelengths \(\lambda_{1}\) and \(\lambda_{2}\) respectively. The maximum kinetic energies of electrons emitted in the two cases are \(E_{1}\) and \(E_{2}\) respectively. The work function of metal is.
357753
The maximum kinetic energy of the photoelectrons depends only on
1 Incident angle
2 Potential
3 Pressure
4 Frequency
Explanation:
using \({E_k} = hv - h{v_0},\) maximum kinetic energy of photoelctrons depends only on frequency.
KCET - 2014
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357754
The correct graph between the maximum energy of a photoelectron and the inverse of wavelength of the incident radiation is given by the curve
1 \(A\)
2 \(B\)
3 \(C\)
4 None of the above
Explanation:
\(K_{\max }=h v-\phi=\dfrac{h c}{\lambda}-\phi\) i.e., graph between \(K_{\max }\) and \(\dfrac{1}{\lambda}\) will be straight line having slope (\(hc\)) and intercept \(\phi\) on -\(KE\) axis. So option (1) is correct.
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357755
Light of energy \({E}\) falls normally on a metal of work function \({E / 3}\). The kinetic energies \({(K)}\) of the photo electrons are
1 \({K=\dfrac{2 E}{3}}\)
2 \({K=\dfrac{E}{3}}\)
3 \({0 \leq K \leq \dfrac{2 E}{3}}\)
4 \({0 \leq K \leq \dfrac{E}{3}}\)
Explanation:
From Einstein's equation for photoelectric emission, maximum kinetic energy of photo electrons is given by \(K{E_{\max }} = E - {\omega _0}\) \( = E - \frac{E}{3} = \frac{{2E}}{3}\) \({\therefore K E}\) lies from 0 to \({\dfrac{2 E}{3}}\). Correct option is (3).
357751
For the photoelectric effect, the maximum kinetic energy \(E_{k}\) of the emitted photoelectrons is plotted against the frequency \(v\) of the incident photons as shown in the figure. The slope of the curve gives
1 Charge of the electron
2 Work function of the metal
3 Planck's constant
4 Ratio of the Planck's constant to electronic charge
Explanation:
Comparing Einstein's equation \(K_{\max }=h v-\phi\), with \(y=m x+c\), we get slope, \(m=h\)
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357752
A photometal is illuminated by lights of wavelengths \(\lambda_{1}\) and \(\lambda_{2}\) respectively. The maximum kinetic energies of electrons emitted in the two cases are \(E_{1}\) and \(E_{2}\) respectively. The work function of metal is.
357753
The maximum kinetic energy of the photoelectrons depends only on
1 Incident angle
2 Potential
3 Pressure
4 Frequency
Explanation:
using \({E_k} = hv - h{v_0},\) maximum kinetic energy of photoelctrons depends only on frequency.
KCET - 2014
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357754
The correct graph between the maximum energy of a photoelectron and the inverse of wavelength of the incident radiation is given by the curve
1 \(A\)
2 \(B\)
3 \(C\)
4 None of the above
Explanation:
\(K_{\max }=h v-\phi=\dfrac{h c}{\lambda}-\phi\) i.e., graph between \(K_{\max }\) and \(\dfrac{1}{\lambda}\) will be straight line having slope (\(hc\)) and intercept \(\phi\) on -\(KE\) axis. So option (1) is correct.
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357755
Light of energy \({E}\) falls normally on a metal of work function \({E / 3}\). The kinetic energies \({(K)}\) of the photo electrons are
1 \({K=\dfrac{2 E}{3}}\)
2 \({K=\dfrac{E}{3}}\)
3 \({0 \leq K \leq \dfrac{2 E}{3}}\)
4 \({0 \leq K \leq \dfrac{E}{3}}\)
Explanation:
From Einstein's equation for photoelectric emission, maximum kinetic energy of photo electrons is given by \(K{E_{\max }} = E - {\omega _0}\) \( = E - \frac{E}{3} = \frac{{2E}}{3}\) \({\therefore K E}\) lies from 0 to \({\dfrac{2 E}{3}}\). Correct option is (3).
357751
For the photoelectric effect, the maximum kinetic energy \(E_{k}\) of the emitted photoelectrons is plotted against the frequency \(v\) of the incident photons as shown in the figure. The slope of the curve gives
1 Charge of the electron
2 Work function of the metal
3 Planck's constant
4 Ratio of the Planck's constant to electronic charge
Explanation:
Comparing Einstein's equation \(K_{\max }=h v-\phi\), with \(y=m x+c\), we get slope, \(m=h\)
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357752
A photometal is illuminated by lights of wavelengths \(\lambda_{1}\) and \(\lambda_{2}\) respectively. The maximum kinetic energies of electrons emitted in the two cases are \(E_{1}\) and \(E_{2}\) respectively. The work function of metal is.
357753
The maximum kinetic energy of the photoelectrons depends only on
1 Incident angle
2 Potential
3 Pressure
4 Frequency
Explanation:
using \({E_k} = hv - h{v_0},\) maximum kinetic energy of photoelctrons depends only on frequency.
KCET - 2014
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357754
The correct graph between the maximum energy of a photoelectron and the inverse of wavelength of the incident radiation is given by the curve
1 \(A\)
2 \(B\)
3 \(C\)
4 None of the above
Explanation:
\(K_{\max }=h v-\phi=\dfrac{h c}{\lambda}-\phi\) i.e., graph between \(K_{\max }\) and \(\dfrac{1}{\lambda}\) will be straight line having slope (\(hc\)) and intercept \(\phi\) on -\(KE\) axis. So option (1) is correct.
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357755
Light of energy \({E}\) falls normally on a metal of work function \({E / 3}\). The kinetic energies \({(K)}\) of the photo electrons are
1 \({K=\dfrac{2 E}{3}}\)
2 \({K=\dfrac{E}{3}}\)
3 \({0 \leq K \leq \dfrac{2 E}{3}}\)
4 \({0 \leq K \leq \dfrac{E}{3}}\)
Explanation:
From Einstein's equation for photoelectric emission, maximum kinetic energy of photo electrons is given by \(K{E_{\max }} = E - {\omega _0}\) \( = E - \frac{E}{3} = \frac{{2E}}{3}\) \({\therefore K E}\) lies from 0 to \({\dfrac{2 E}{3}}\). Correct option is (3).
