NEET Test Series from KOTA - 10 Papers In MS WORD
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Dual nature of radiation and Matter
142320
The threshold frequency for photoelectric effect on sodium corresponds to a wavelength of $5000 \AA$. Its work function is
1 $4 \times 10^{-19} \mathrm{~J}$
2 $1 \mathrm{~J}$
3 $2 \times 10^{-19} \mathrm{~J}$
4 $13 \times 10^{-19} \mathrm{~J}$
Explanation:
A Given that, $\lambda_{\mathrm{o}}=5000 \AA=5000 \times 10^{-10} \mathrm{~m}$ $\mathrm{c}=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$ We know that, Work function $(\phi)=\frac{h c}{\lambda_{o}}$ Putting these value, we get - $\phi=\frac{6.62 \times 10^{-34} \times 3 \times 10^{8}}{5000 \times 10^{-10}}$ $\phi \square 4 \times 10^{-19} \mathrm{~J}$
AIPMT - 1988
Dual nature of radiation and Matter
142321
If the threshold wavelength for a certain metal is $2000 \AA$, then the work function of the metal is
1 $6.2 \mathrm{~J}$
2 $6.2 \mathrm{eV}$
3 $6.2 \mathrm{MeV}$
4 $6.2 \mathrm{keV}$
Explanation:
B Given that, $\lambda=2000 \AA=2 \times 10^{-7} \mathrm{~m}$ $\mathrm{hc}=1240 \mathrm{eV} \mathrm{nm}$ We know that, $\phi=\frac{\mathrm{hc}}{\lambda}$ Putting these value, we get - $\phi =\frac{1240}{2 \times 10^{-7}}$ $\phi =9.9 \times 10^{-19} \mathrm{~J}$ $\phi =\frac{9.9 \times 10^{-19}}{1.6 \times 10^{-19}} \mathrm{eV}$ $\phi =6.2 \mathrm{eV}$
EAMCET-1995
Dual nature of radiation and Matter
142323
Einstein's work on photoelectric effect gives support to
1 $\mathrm{E}=m \mathrm{c}^{2}$
2 $\mathrm{E}=\mathrm{hv}$
3 $\mathrm{hv}=\frac{1}{2} \mathrm{mv}^{2}$
4 $\mathrm{E}=\frac{\mathrm{h}}{\lambda}$
Explanation:
B According to Einstein's work on photoelectric effect- $\mathrm{h} v=\phi_{\mathrm{o}}+\frac{1}{2} \mathrm{mv}^{2}$ $\mathrm{~h} v=\mathrm{h} v_{o}+\frac{1}{2} \mathrm{mv}^{2}$ Planck equation, $\mathrm{E}=\mathrm{h} v$ Where, $v=$ Frequency $\mathrm{h}=\text { Planck constant }$
AIPMT - 2000
Dual nature of radiation and Matter
142324
In photoelectric emission process from a metal of work function $1.8 \mathrm{eV}$, the kinetic energy of most energetic electrons is $0.5 \mathrm{eV}$. The corresponding stopping potential is
1 $1.3 \mathrm{~V}$
2 $0.5 \mathrm{~V}$
3 $2.3 \mathrm{~V}$
4 $1.8 \mathrm{~V}$
Explanation:
B Given that, $\phi_{\mathrm{o}}=1.8 \mathrm{eV}$, kinetic energy of most energetic electron $=0.5 \mathrm{eV}$ We know that, stopping potential is equal to the maximum kinetic energy of electron, K.E. = P.E. $0.5 \mathrm{eV}=\mathrm{eV}_{\mathrm{o}}$ $\mathrm{V}_{\mathrm{o}}=0.5 \mathrm{~V}$ Where, $\mathrm{V}_{\mathrm{o}}=$ Stopping potential
142320
The threshold frequency for photoelectric effect on sodium corresponds to a wavelength of $5000 \AA$. Its work function is
1 $4 \times 10^{-19} \mathrm{~J}$
2 $1 \mathrm{~J}$
3 $2 \times 10^{-19} \mathrm{~J}$
4 $13 \times 10^{-19} \mathrm{~J}$
Explanation:
A Given that, $\lambda_{\mathrm{o}}=5000 \AA=5000 \times 10^{-10} \mathrm{~m}$ $\mathrm{c}=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$ We know that, Work function $(\phi)=\frac{h c}{\lambda_{o}}$ Putting these value, we get - $\phi=\frac{6.62 \times 10^{-34} \times 3 \times 10^{8}}{5000 \times 10^{-10}}$ $\phi \square 4 \times 10^{-19} \mathrm{~J}$
AIPMT - 1988
Dual nature of radiation and Matter
142321
If the threshold wavelength for a certain metal is $2000 \AA$, then the work function of the metal is
1 $6.2 \mathrm{~J}$
2 $6.2 \mathrm{eV}$
3 $6.2 \mathrm{MeV}$
4 $6.2 \mathrm{keV}$
Explanation:
B Given that, $\lambda=2000 \AA=2 \times 10^{-7} \mathrm{~m}$ $\mathrm{hc}=1240 \mathrm{eV} \mathrm{nm}$ We know that, $\phi=\frac{\mathrm{hc}}{\lambda}$ Putting these value, we get - $\phi =\frac{1240}{2 \times 10^{-7}}$ $\phi =9.9 \times 10^{-19} \mathrm{~J}$ $\phi =\frac{9.9 \times 10^{-19}}{1.6 \times 10^{-19}} \mathrm{eV}$ $\phi =6.2 \mathrm{eV}$
EAMCET-1995
Dual nature of radiation and Matter
142323
Einstein's work on photoelectric effect gives support to
1 $\mathrm{E}=m \mathrm{c}^{2}$
2 $\mathrm{E}=\mathrm{hv}$
3 $\mathrm{hv}=\frac{1}{2} \mathrm{mv}^{2}$
4 $\mathrm{E}=\frac{\mathrm{h}}{\lambda}$
Explanation:
B According to Einstein's work on photoelectric effect- $\mathrm{h} v=\phi_{\mathrm{o}}+\frac{1}{2} \mathrm{mv}^{2}$ $\mathrm{~h} v=\mathrm{h} v_{o}+\frac{1}{2} \mathrm{mv}^{2}$ Planck equation, $\mathrm{E}=\mathrm{h} v$ Where, $v=$ Frequency $\mathrm{h}=\text { Planck constant }$
AIPMT - 2000
Dual nature of radiation and Matter
142324
In photoelectric emission process from a metal of work function $1.8 \mathrm{eV}$, the kinetic energy of most energetic electrons is $0.5 \mathrm{eV}$. The corresponding stopping potential is
1 $1.3 \mathrm{~V}$
2 $0.5 \mathrm{~V}$
3 $2.3 \mathrm{~V}$
4 $1.8 \mathrm{~V}$
Explanation:
B Given that, $\phi_{\mathrm{o}}=1.8 \mathrm{eV}$, kinetic energy of most energetic electron $=0.5 \mathrm{eV}$ We know that, stopping potential is equal to the maximum kinetic energy of electron, K.E. = P.E. $0.5 \mathrm{eV}=\mathrm{eV}_{\mathrm{o}}$ $\mathrm{V}_{\mathrm{o}}=0.5 \mathrm{~V}$ Where, $\mathrm{V}_{\mathrm{o}}=$ Stopping potential
142320
The threshold frequency for photoelectric effect on sodium corresponds to a wavelength of $5000 \AA$. Its work function is
1 $4 \times 10^{-19} \mathrm{~J}$
2 $1 \mathrm{~J}$
3 $2 \times 10^{-19} \mathrm{~J}$
4 $13 \times 10^{-19} \mathrm{~J}$
Explanation:
A Given that, $\lambda_{\mathrm{o}}=5000 \AA=5000 \times 10^{-10} \mathrm{~m}$ $\mathrm{c}=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$ We know that, Work function $(\phi)=\frac{h c}{\lambda_{o}}$ Putting these value, we get - $\phi=\frac{6.62 \times 10^{-34} \times 3 \times 10^{8}}{5000 \times 10^{-10}}$ $\phi \square 4 \times 10^{-19} \mathrm{~J}$
AIPMT - 1988
Dual nature of radiation and Matter
142321
If the threshold wavelength for a certain metal is $2000 \AA$, then the work function of the metal is
1 $6.2 \mathrm{~J}$
2 $6.2 \mathrm{eV}$
3 $6.2 \mathrm{MeV}$
4 $6.2 \mathrm{keV}$
Explanation:
B Given that, $\lambda=2000 \AA=2 \times 10^{-7} \mathrm{~m}$ $\mathrm{hc}=1240 \mathrm{eV} \mathrm{nm}$ We know that, $\phi=\frac{\mathrm{hc}}{\lambda}$ Putting these value, we get - $\phi =\frac{1240}{2 \times 10^{-7}}$ $\phi =9.9 \times 10^{-19} \mathrm{~J}$ $\phi =\frac{9.9 \times 10^{-19}}{1.6 \times 10^{-19}} \mathrm{eV}$ $\phi =6.2 \mathrm{eV}$
EAMCET-1995
Dual nature of radiation and Matter
142323
Einstein's work on photoelectric effect gives support to
1 $\mathrm{E}=m \mathrm{c}^{2}$
2 $\mathrm{E}=\mathrm{hv}$
3 $\mathrm{hv}=\frac{1}{2} \mathrm{mv}^{2}$
4 $\mathrm{E}=\frac{\mathrm{h}}{\lambda}$
Explanation:
B According to Einstein's work on photoelectric effect- $\mathrm{h} v=\phi_{\mathrm{o}}+\frac{1}{2} \mathrm{mv}^{2}$ $\mathrm{~h} v=\mathrm{h} v_{o}+\frac{1}{2} \mathrm{mv}^{2}$ Planck equation, $\mathrm{E}=\mathrm{h} v$ Where, $v=$ Frequency $\mathrm{h}=\text { Planck constant }$
AIPMT - 2000
Dual nature of radiation and Matter
142324
In photoelectric emission process from a metal of work function $1.8 \mathrm{eV}$, the kinetic energy of most energetic electrons is $0.5 \mathrm{eV}$. The corresponding stopping potential is
1 $1.3 \mathrm{~V}$
2 $0.5 \mathrm{~V}$
3 $2.3 \mathrm{~V}$
4 $1.8 \mathrm{~V}$
Explanation:
B Given that, $\phi_{\mathrm{o}}=1.8 \mathrm{eV}$, kinetic energy of most energetic electron $=0.5 \mathrm{eV}$ We know that, stopping potential is equal to the maximum kinetic energy of electron, K.E. = P.E. $0.5 \mathrm{eV}=\mathrm{eV}_{\mathrm{o}}$ $\mathrm{V}_{\mathrm{o}}=0.5 \mathrm{~V}$ Where, $\mathrm{V}_{\mathrm{o}}=$ Stopping potential
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Dual nature of radiation and Matter
142320
The threshold frequency for photoelectric effect on sodium corresponds to a wavelength of $5000 \AA$. Its work function is
1 $4 \times 10^{-19} \mathrm{~J}$
2 $1 \mathrm{~J}$
3 $2 \times 10^{-19} \mathrm{~J}$
4 $13 \times 10^{-19} \mathrm{~J}$
Explanation:
A Given that, $\lambda_{\mathrm{o}}=5000 \AA=5000 \times 10^{-10} \mathrm{~m}$ $\mathrm{c}=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$ We know that, Work function $(\phi)=\frac{h c}{\lambda_{o}}$ Putting these value, we get - $\phi=\frac{6.62 \times 10^{-34} \times 3 \times 10^{8}}{5000 \times 10^{-10}}$ $\phi \square 4 \times 10^{-19} \mathrm{~J}$
AIPMT - 1988
Dual nature of radiation and Matter
142321
If the threshold wavelength for a certain metal is $2000 \AA$, then the work function of the metal is
1 $6.2 \mathrm{~J}$
2 $6.2 \mathrm{eV}$
3 $6.2 \mathrm{MeV}$
4 $6.2 \mathrm{keV}$
Explanation:
B Given that, $\lambda=2000 \AA=2 \times 10^{-7} \mathrm{~m}$ $\mathrm{hc}=1240 \mathrm{eV} \mathrm{nm}$ We know that, $\phi=\frac{\mathrm{hc}}{\lambda}$ Putting these value, we get - $\phi =\frac{1240}{2 \times 10^{-7}}$ $\phi =9.9 \times 10^{-19} \mathrm{~J}$ $\phi =\frac{9.9 \times 10^{-19}}{1.6 \times 10^{-19}} \mathrm{eV}$ $\phi =6.2 \mathrm{eV}$
EAMCET-1995
Dual nature of radiation and Matter
142323
Einstein's work on photoelectric effect gives support to
1 $\mathrm{E}=m \mathrm{c}^{2}$
2 $\mathrm{E}=\mathrm{hv}$
3 $\mathrm{hv}=\frac{1}{2} \mathrm{mv}^{2}$
4 $\mathrm{E}=\frac{\mathrm{h}}{\lambda}$
Explanation:
B According to Einstein's work on photoelectric effect- $\mathrm{h} v=\phi_{\mathrm{o}}+\frac{1}{2} \mathrm{mv}^{2}$ $\mathrm{~h} v=\mathrm{h} v_{o}+\frac{1}{2} \mathrm{mv}^{2}$ Planck equation, $\mathrm{E}=\mathrm{h} v$ Where, $v=$ Frequency $\mathrm{h}=\text { Planck constant }$
AIPMT - 2000
Dual nature of radiation and Matter
142324
In photoelectric emission process from a metal of work function $1.8 \mathrm{eV}$, the kinetic energy of most energetic electrons is $0.5 \mathrm{eV}$. The corresponding stopping potential is
1 $1.3 \mathrm{~V}$
2 $0.5 \mathrm{~V}$
3 $2.3 \mathrm{~V}$
4 $1.8 \mathrm{~V}$
Explanation:
B Given that, $\phi_{\mathrm{o}}=1.8 \mathrm{eV}$, kinetic energy of most energetic electron $=0.5 \mathrm{eV}$ We know that, stopping potential is equal to the maximum kinetic energy of electron, K.E. = P.E. $0.5 \mathrm{eV}=\mathrm{eV}_{\mathrm{o}}$ $\mathrm{V}_{\mathrm{o}}=0.5 \mathrm{~V}$ Where, $\mathrm{V}_{\mathrm{o}}=$ Stopping potential
