142293
The threshold wavelength for photoelectric effect of a metal is 6500 . The work function of the metal is approximately
1 $2 \mathrm{eV}$
2 $1 \mathrm{eV}$
3 $0.1 \mathrm{eV}$
4 $3 \mathrm{eV}$
Explanation:
A Given that, $\lambda_{\mathrm{o}}=6500 \AA$ Work function $\left(\phi_{\mathrm{o}}\right)=\frac{\mathrm{hc}}{\lambda_{\mathrm{o}}}$ $\phi_{\mathrm{o}} =\frac{12375}{6500} \quad[\because \mathrm{hc}=12375]$ $=1.903 \mathrm{eV}$ $\phi_{\mathrm{o}} \sqcup 2 \mathrm{eV}$ Hence, the work function of metal is $2 \mathrm{eV}$.
J and K CET- 1999
Dual nature of radiation and Matter
142294
The threshold wavelength is $2000 \AA$. The work function is
142296
If an electron, a proton and an alpha particle have same de Broglie wavelengths then which statement about their kinetic energies is correct?
1 Kinetic energy of alpha particle is least
2 Kinetic energy of proton is least
3 Kinetic energy of electron is least
4 Kinetic energies of all three are same
Explanation:
A Given, Let the wavelength of electron, proton and alpha particle are $\lambda_{\mathrm{e}}, \lambda_{\mathrm{p}}$ and $\lambda_{\alpha}$ respectively- $\lambda_{\mathrm{e}}=\lambda_{\mathrm{p}}=\lambda_{\alpha}$ Since, $\quad \lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}}}$ Then, $\quad \frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\mathrm{e}} \mathrm{K}_{\mathrm{e}}}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\mathrm{p}} \mathrm{K}_{\mathrm{p}}}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\alpha} \mathrm{K}_{\alpha}}}$ $\mathrm{m}_{\mathrm{e}} \mathrm{K}_{\mathrm{e}}=\mathrm{m}_{\mathrm{p}} \mathrm{K}_{\mathrm{p}}=\mathrm{m}_{\alpha} \mathrm{K}_{\alpha}$ $\mathrm{m}_{\mathrm{e}} \lt \mathrm{m}_{\mathrm{p}} \lt \mathrm{m}_{\alpha}$ $\mathrm{K}_{\mathrm{e}}>\mathrm{K}_{\mathrm{p}}>\mathrm{K}_{\alpha}$ Hence, the kinetic energy of alpha particle is least.
142293
The threshold wavelength for photoelectric effect of a metal is 6500 . The work function of the metal is approximately
1 $2 \mathrm{eV}$
2 $1 \mathrm{eV}$
3 $0.1 \mathrm{eV}$
4 $3 \mathrm{eV}$
Explanation:
A Given that, $\lambda_{\mathrm{o}}=6500 \AA$ Work function $\left(\phi_{\mathrm{o}}\right)=\frac{\mathrm{hc}}{\lambda_{\mathrm{o}}}$ $\phi_{\mathrm{o}} =\frac{12375}{6500} \quad[\because \mathrm{hc}=12375]$ $=1.903 \mathrm{eV}$ $\phi_{\mathrm{o}} \sqcup 2 \mathrm{eV}$ Hence, the work function of metal is $2 \mathrm{eV}$.
J and K CET- 1999
Dual nature of radiation and Matter
142294
The threshold wavelength is $2000 \AA$. The work function is
142296
If an electron, a proton and an alpha particle have same de Broglie wavelengths then which statement about their kinetic energies is correct?
1 Kinetic energy of alpha particle is least
2 Kinetic energy of proton is least
3 Kinetic energy of electron is least
4 Kinetic energies of all three are same
Explanation:
A Given, Let the wavelength of electron, proton and alpha particle are $\lambda_{\mathrm{e}}, \lambda_{\mathrm{p}}$ and $\lambda_{\alpha}$ respectively- $\lambda_{\mathrm{e}}=\lambda_{\mathrm{p}}=\lambda_{\alpha}$ Since, $\quad \lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}}}$ Then, $\quad \frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\mathrm{e}} \mathrm{K}_{\mathrm{e}}}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\mathrm{p}} \mathrm{K}_{\mathrm{p}}}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\alpha} \mathrm{K}_{\alpha}}}$ $\mathrm{m}_{\mathrm{e}} \mathrm{K}_{\mathrm{e}}=\mathrm{m}_{\mathrm{p}} \mathrm{K}_{\mathrm{p}}=\mathrm{m}_{\alpha} \mathrm{K}_{\alpha}$ $\mathrm{m}_{\mathrm{e}} \lt \mathrm{m}_{\mathrm{p}} \lt \mathrm{m}_{\alpha}$ $\mathrm{K}_{\mathrm{e}}>\mathrm{K}_{\mathrm{p}}>\mathrm{K}_{\alpha}$ Hence, the kinetic energy of alpha particle is least.
142293
The threshold wavelength for photoelectric effect of a metal is 6500 . The work function of the metal is approximately
1 $2 \mathrm{eV}$
2 $1 \mathrm{eV}$
3 $0.1 \mathrm{eV}$
4 $3 \mathrm{eV}$
Explanation:
A Given that, $\lambda_{\mathrm{o}}=6500 \AA$ Work function $\left(\phi_{\mathrm{o}}\right)=\frac{\mathrm{hc}}{\lambda_{\mathrm{o}}}$ $\phi_{\mathrm{o}} =\frac{12375}{6500} \quad[\because \mathrm{hc}=12375]$ $=1.903 \mathrm{eV}$ $\phi_{\mathrm{o}} \sqcup 2 \mathrm{eV}$ Hence, the work function of metal is $2 \mathrm{eV}$.
J and K CET- 1999
Dual nature of radiation and Matter
142294
The threshold wavelength is $2000 \AA$. The work function is
142296
If an electron, a proton and an alpha particle have same de Broglie wavelengths then which statement about their kinetic energies is correct?
1 Kinetic energy of alpha particle is least
2 Kinetic energy of proton is least
3 Kinetic energy of electron is least
4 Kinetic energies of all three are same
Explanation:
A Given, Let the wavelength of electron, proton and alpha particle are $\lambda_{\mathrm{e}}, \lambda_{\mathrm{p}}$ and $\lambda_{\alpha}$ respectively- $\lambda_{\mathrm{e}}=\lambda_{\mathrm{p}}=\lambda_{\alpha}$ Since, $\quad \lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}}}$ Then, $\quad \frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\mathrm{e}} \mathrm{K}_{\mathrm{e}}}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\mathrm{p}} \mathrm{K}_{\mathrm{p}}}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\alpha} \mathrm{K}_{\alpha}}}$ $\mathrm{m}_{\mathrm{e}} \mathrm{K}_{\mathrm{e}}=\mathrm{m}_{\mathrm{p}} \mathrm{K}_{\mathrm{p}}=\mathrm{m}_{\alpha} \mathrm{K}_{\alpha}$ $\mathrm{m}_{\mathrm{e}} \lt \mathrm{m}_{\mathrm{p}} \lt \mathrm{m}_{\alpha}$ $\mathrm{K}_{\mathrm{e}}>\mathrm{K}_{\mathrm{p}}>\mathrm{K}_{\alpha}$ Hence, the kinetic energy of alpha particle is least.
142293
The threshold wavelength for photoelectric effect of a metal is 6500 . The work function of the metal is approximately
1 $2 \mathrm{eV}$
2 $1 \mathrm{eV}$
3 $0.1 \mathrm{eV}$
4 $3 \mathrm{eV}$
Explanation:
A Given that, $\lambda_{\mathrm{o}}=6500 \AA$ Work function $\left(\phi_{\mathrm{o}}\right)=\frac{\mathrm{hc}}{\lambda_{\mathrm{o}}}$ $\phi_{\mathrm{o}} =\frac{12375}{6500} \quad[\because \mathrm{hc}=12375]$ $=1.903 \mathrm{eV}$ $\phi_{\mathrm{o}} \sqcup 2 \mathrm{eV}$ Hence, the work function of metal is $2 \mathrm{eV}$.
J and K CET- 1999
Dual nature of radiation and Matter
142294
The threshold wavelength is $2000 \AA$. The work function is
142296
If an electron, a proton and an alpha particle have same de Broglie wavelengths then which statement about their kinetic energies is correct?
1 Kinetic energy of alpha particle is least
2 Kinetic energy of proton is least
3 Kinetic energy of electron is least
4 Kinetic energies of all three are same
Explanation:
A Given, Let the wavelength of electron, proton and alpha particle are $\lambda_{\mathrm{e}}, \lambda_{\mathrm{p}}$ and $\lambda_{\alpha}$ respectively- $\lambda_{\mathrm{e}}=\lambda_{\mathrm{p}}=\lambda_{\alpha}$ Since, $\quad \lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}}}$ Then, $\quad \frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\mathrm{e}} \mathrm{K}_{\mathrm{e}}}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\mathrm{p}} \mathrm{K}_{\mathrm{p}}}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\alpha} \mathrm{K}_{\alpha}}}$ $\mathrm{m}_{\mathrm{e}} \mathrm{K}_{\mathrm{e}}=\mathrm{m}_{\mathrm{p}} \mathrm{K}_{\mathrm{p}}=\mathrm{m}_{\alpha} \mathrm{K}_{\alpha}$ $\mathrm{m}_{\mathrm{e}} \lt \mathrm{m}_{\mathrm{p}} \lt \mathrm{m}_{\alpha}$ $\mathrm{K}_{\mathrm{e}}>\mathrm{K}_{\mathrm{p}}>\mathrm{K}_{\alpha}$ Hence, the kinetic energy of alpha particle is least.
142293
The threshold wavelength for photoelectric effect of a metal is 6500 . The work function of the metal is approximately
1 $2 \mathrm{eV}$
2 $1 \mathrm{eV}$
3 $0.1 \mathrm{eV}$
4 $3 \mathrm{eV}$
Explanation:
A Given that, $\lambda_{\mathrm{o}}=6500 \AA$ Work function $\left(\phi_{\mathrm{o}}\right)=\frac{\mathrm{hc}}{\lambda_{\mathrm{o}}}$ $\phi_{\mathrm{o}} =\frac{12375}{6500} \quad[\because \mathrm{hc}=12375]$ $=1.903 \mathrm{eV}$ $\phi_{\mathrm{o}} \sqcup 2 \mathrm{eV}$ Hence, the work function of metal is $2 \mathrm{eV}$.
J and K CET- 1999
Dual nature of radiation and Matter
142294
The threshold wavelength is $2000 \AA$. The work function is
142296
If an electron, a proton and an alpha particle have same de Broglie wavelengths then which statement about their kinetic energies is correct?
1 Kinetic energy of alpha particle is least
2 Kinetic energy of proton is least
3 Kinetic energy of electron is least
4 Kinetic energies of all three are same
Explanation:
A Given, Let the wavelength of electron, proton and alpha particle are $\lambda_{\mathrm{e}}, \lambda_{\mathrm{p}}$ and $\lambda_{\alpha}$ respectively- $\lambda_{\mathrm{e}}=\lambda_{\mathrm{p}}=\lambda_{\alpha}$ Since, $\quad \lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}}}$ Then, $\quad \frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\mathrm{e}} \mathrm{K}_{\mathrm{e}}}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\mathrm{p}} \mathrm{K}_{\mathrm{p}}}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\alpha} \mathrm{K}_{\alpha}}}$ $\mathrm{m}_{\mathrm{e}} \mathrm{K}_{\mathrm{e}}=\mathrm{m}_{\mathrm{p}} \mathrm{K}_{\mathrm{p}}=\mathrm{m}_{\alpha} \mathrm{K}_{\alpha}$ $\mathrm{m}_{\mathrm{e}} \lt \mathrm{m}_{\mathrm{p}} \lt \mathrm{m}_{\alpha}$ $\mathrm{K}_{\mathrm{e}}>\mathrm{K}_{\mathrm{p}}>\mathrm{K}_{\alpha}$ Hence, the kinetic energy of alpha particle is least.