142214
The retarding potential necessary to stop the emission of photoelectrons, when a target material of work function $1.24 \mathrm{eV}$ is irradiated with light of wavelength $4.35 \times 10^{-7} \mathrm{~m}$ is
1 $4.09 \mathrm{eV}$
2 $2.84 \mathrm{eV}$
3 $1.60 \mathrm{eV}$
4 $0.36 \mathrm{eV}$
Explanation:
C Given that, Light of wavelength $(\lambda)=4.35 \times 10^{-7} \mathrm{~m}$ Work function $(\phi)=1.24 \mathrm{eV}$ Einstein's photoelectric equation- $\mathrm{eV}_{\mathrm{o}}=\frac{\mathrm{hc}}{\lambda}-\phi \quad\left(\because \mathrm{c}=3 \times 10^{8} \mathrm{~m} / \mathrm{s}\right)$ $\mathrm{eV}_{\mathrm{o}}=\frac{6.62 \times 10^{-34} \times 3 \times 10^{8}}{4.36 \times 10^{-7}}-1.24$ $\mathrm{~V}_{\mathrm{o}}=\frac{6.62 \times 10^{-34} \times 3 \times 10^{8}}{1.6 \times 10^{-19} \times 4.36 \times 10^{-7}}-1.24$ $\mathrm{~V}_{\mathrm{o}}=2.84-1.24$ $\mathrm{~V}_{\mathrm{o}}=1.60 \mathrm{eV}$
AP EAMCET (18.09.2020) Shift-I
Dual nature of radiation and Matter
142215
Radiation of two photons having energies twice and five times the work function of a metal are incident successively on the metal surface. Find out the ratio of maximum velocity of photo electrons emitted in the two cases.
1 $\mathrm{v}_{1} / \mathrm{v}_{2}=1 / 3$
2 $\mathrm{v}_{1} / \mathrm{v}_{2}=1 / 4$
3 $\mathrm{v}_{1} / \mathrm{v}_{2}=1$
4 $\mathrm{v}_{1} / \mathrm{v}_{2}=1 / 2$
Explanation:
D Given that, $\mathrm{E}_{1}=2 \phi$ $\mathrm{E}_{2}=5 \phi$ $\because \quad \mathrm{E}_{1}=\phi+\mathrm{K} . \mathrm{E}_{1}$ Putting the value of $E_{1}$, we get - $2 \phi-\phi=K . E_{1}$ $\phi=K . E_{1}$ $\text { And } \quad \mathrm{E}_{2}=\phi+\mathrm{KE}_{2}$ Putting the value of $\mathrm{E}_{1}$, we get - $5 \phi=\phi+\mathrm{KE}_{2}$ $\mathrm{KE}_{2}=4 \phi$ Dividing equation (i) by (ii), we get- $\therefore \quad \frac{\frac{1}{2} \mathrm{mv}_{1}^{2}}{\frac{1}{2} \mathrm{mv}_{2}^{2}}=\frac{1}{4}$ $\frac{v_{1}^{2}}{v_{2}^{2}}=\frac{1}{4}$ $\frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}=\frac{1}{2}$
SRMJEE-2019
Dual nature of radiation and Matter
142217
If frequency of a photon is $6 \times 10^{14} \mathrm{~Hz}$, then find its wavelength [ Take, speed of light, $\mathrm{c}=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$ ]
1 $500 \AA$
2 $500 \mathrm{~nm}$
3 $200 \AA$
4 $200 \mathrm{~nm}$
Explanation:
B Given that, $\mathrm{c}=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$ Frequency of photon $(v)=6 \times 10^{14} \mathrm{~Hz}, \lambda=?$ We know that, \(v=\frac{\mathrm{c}}{\lambda}\) Putting these value, we get- \(6 \times 10^{14}=\frac{3 \times 10^{8}}{\lambda}\) $\quad \lambda=0.5 \times 10^{-6}$ $\lambda=5 \times 10^{-7} \times \frac{100}{100}$ $\lambda=5 \times 100 \times 10^{-9} \mathrm{~m} \quad\left[\because 10^{-9} \mathrm{~m}=1 \mathrm{~nm}\right]$ $\lambda=500 \mathrm{~nm}$
142214
The retarding potential necessary to stop the emission of photoelectrons, when a target material of work function $1.24 \mathrm{eV}$ is irradiated with light of wavelength $4.35 \times 10^{-7} \mathrm{~m}$ is
1 $4.09 \mathrm{eV}$
2 $2.84 \mathrm{eV}$
3 $1.60 \mathrm{eV}$
4 $0.36 \mathrm{eV}$
Explanation:
C Given that, Light of wavelength $(\lambda)=4.35 \times 10^{-7} \mathrm{~m}$ Work function $(\phi)=1.24 \mathrm{eV}$ Einstein's photoelectric equation- $\mathrm{eV}_{\mathrm{o}}=\frac{\mathrm{hc}}{\lambda}-\phi \quad\left(\because \mathrm{c}=3 \times 10^{8} \mathrm{~m} / \mathrm{s}\right)$ $\mathrm{eV}_{\mathrm{o}}=\frac{6.62 \times 10^{-34} \times 3 \times 10^{8}}{4.36 \times 10^{-7}}-1.24$ $\mathrm{~V}_{\mathrm{o}}=\frac{6.62 \times 10^{-34} \times 3 \times 10^{8}}{1.6 \times 10^{-19} \times 4.36 \times 10^{-7}}-1.24$ $\mathrm{~V}_{\mathrm{o}}=2.84-1.24$ $\mathrm{~V}_{\mathrm{o}}=1.60 \mathrm{eV}$
AP EAMCET (18.09.2020) Shift-I
Dual nature of radiation and Matter
142215
Radiation of two photons having energies twice and five times the work function of a metal are incident successively on the metal surface. Find out the ratio of maximum velocity of photo electrons emitted in the two cases.
1 $\mathrm{v}_{1} / \mathrm{v}_{2}=1 / 3$
2 $\mathrm{v}_{1} / \mathrm{v}_{2}=1 / 4$
3 $\mathrm{v}_{1} / \mathrm{v}_{2}=1$
4 $\mathrm{v}_{1} / \mathrm{v}_{2}=1 / 2$
Explanation:
D Given that, $\mathrm{E}_{1}=2 \phi$ $\mathrm{E}_{2}=5 \phi$ $\because \quad \mathrm{E}_{1}=\phi+\mathrm{K} . \mathrm{E}_{1}$ Putting the value of $E_{1}$, we get - $2 \phi-\phi=K . E_{1}$ $\phi=K . E_{1}$ $\text { And } \quad \mathrm{E}_{2}=\phi+\mathrm{KE}_{2}$ Putting the value of $\mathrm{E}_{1}$, we get - $5 \phi=\phi+\mathrm{KE}_{2}$ $\mathrm{KE}_{2}=4 \phi$ Dividing equation (i) by (ii), we get- $\therefore \quad \frac{\frac{1}{2} \mathrm{mv}_{1}^{2}}{\frac{1}{2} \mathrm{mv}_{2}^{2}}=\frac{1}{4}$ $\frac{v_{1}^{2}}{v_{2}^{2}}=\frac{1}{4}$ $\frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}=\frac{1}{2}$
SRMJEE-2019
Dual nature of radiation and Matter
142217
If frequency of a photon is $6 \times 10^{14} \mathrm{~Hz}$, then find its wavelength [ Take, speed of light, $\mathrm{c}=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$ ]
1 $500 \AA$
2 $500 \mathrm{~nm}$
3 $200 \AA$
4 $200 \mathrm{~nm}$
Explanation:
B Given that, $\mathrm{c}=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$ Frequency of photon $(v)=6 \times 10^{14} \mathrm{~Hz}, \lambda=?$ We know that, \(v=\frac{\mathrm{c}}{\lambda}\) Putting these value, we get- \(6 \times 10^{14}=\frac{3 \times 10^{8}}{\lambda}\) $\quad \lambda=0.5 \times 10^{-6}$ $\lambda=5 \times 10^{-7} \times \frac{100}{100}$ $\lambda=5 \times 100 \times 10^{-9} \mathrm{~m} \quad\left[\because 10^{-9} \mathrm{~m}=1 \mathrm{~nm}\right]$ $\lambda=500 \mathrm{~nm}$
142214
The retarding potential necessary to stop the emission of photoelectrons, when a target material of work function $1.24 \mathrm{eV}$ is irradiated with light of wavelength $4.35 \times 10^{-7} \mathrm{~m}$ is
1 $4.09 \mathrm{eV}$
2 $2.84 \mathrm{eV}$
3 $1.60 \mathrm{eV}$
4 $0.36 \mathrm{eV}$
Explanation:
C Given that, Light of wavelength $(\lambda)=4.35 \times 10^{-7} \mathrm{~m}$ Work function $(\phi)=1.24 \mathrm{eV}$ Einstein's photoelectric equation- $\mathrm{eV}_{\mathrm{o}}=\frac{\mathrm{hc}}{\lambda}-\phi \quad\left(\because \mathrm{c}=3 \times 10^{8} \mathrm{~m} / \mathrm{s}\right)$ $\mathrm{eV}_{\mathrm{o}}=\frac{6.62 \times 10^{-34} \times 3 \times 10^{8}}{4.36 \times 10^{-7}}-1.24$ $\mathrm{~V}_{\mathrm{o}}=\frac{6.62 \times 10^{-34} \times 3 \times 10^{8}}{1.6 \times 10^{-19} \times 4.36 \times 10^{-7}}-1.24$ $\mathrm{~V}_{\mathrm{o}}=2.84-1.24$ $\mathrm{~V}_{\mathrm{o}}=1.60 \mathrm{eV}$
AP EAMCET (18.09.2020) Shift-I
Dual nature of radiation and Matter
142215
Radiation of two photons having energies twice and five times the work function of a metal are incident successively on the metal surface. Find out the ratio of maximum velocity of photo electrons emitted in the two cases.
1 $\mathrm{v}_{1} / \mathrm{v}_{2}=1 / 3$
2 $\mathrm{v}_{1} / \mathrm{v}_{2}=1 / 4$
3 $\mathrm{v}_{1} / \mathrm{v}_{2}=1$
4 $\mathrm{v}_{1} / \mathrm{v}_{2}=1 / 2$
Explanation:
D Given that, $\mathrm{E}_{1}=2 \phi$ $\mathrm{E}_{2}=5 \phi$ $\because \quad \mathrm{E}_{1}=\phi+\mathrm{K} . \mathrm{E}_{1}$ Putting the value of $E_{1}$, we get - $2 \phi-\phi=K . E_{1}$ $\phi=K . E_{1}$ $\text { And } \quad \mathrm{E}_{2}=\phi+\mathrm{KE}_{2}$ Putting the value of $\mathrm{E}_{1}$, we get - $5 \phi=\phi+\mathrm{KE}_{2}$ $\mathrm{KE}_{2}=4 \phi$ Dividing equation (i) by (ii), we get- $\therefore \quad \frac{\frac{1}{2} \mathrm{mv}_{1}^{2}}{\frac{1}{2} \mathrm{mv}_{2}^{2}}=\frac{1}{4}$ $\frac{v_{1}^{2}}{v_{2}^{2}}=\frac{1}{4}$ $\frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}=\frac{1}{2}$
SRMJEE-2019
Dual nature of radiation and Matter
142217
If frequency of a photon is $6 \times 10^{14} \mathrm{~Hz}$, then find its wavelength [ Take, speed of light, $\mathrm{c}=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$ ]
1 $500 \AA$
2 $500 \mathrm{~nm}$
3 $200 \AA$
4 $200 \mathrm{~nm}$
Explanation:
B Given that, $\mathrm{c}=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$ Frequency of photon $(v)=6 \times 10^{14} \mathrm{~Hz}, \lambda=?$ We know that, \(v=\frac{\mathrm{c}}{\lambda}\) Putting these value, we get- \(6 \times 10^{14}=\frac{3 \times 10^{8}}{\lambda}\) $\quad \lambda=0.5 \times 10^{-6}$ $\lambda=5 \times 10^{-7} \times \frac{100}{100}$ $\lambda=5 \times 100 \times 10^{-9} \mathrm{~m} \quad\left[\because 10^{-9} \mathrm{~m}=1 \mathrm{~nm}\right]$ $\lambda=500 \mathrm{~nm}$
142214
The retarding potential necessary to stop the emission of photoelectrons, when a target material of work function $1.24 \mathrm{eV}$ is irradiated with light of wavelength $4.35 \times 10^{-7} \mathrm{~m}$ is
1 $4.09 \mathrm{eV}$
2 $2.84 \mathrm{eV}$
3 $1.60 \mathrm{eV}$
4 $0.36 \mathrm{eV}$
Explanation:
C Given that, Light of wavelength $(\lambda)=4.35 \times 10^{-7} \mathrm{~m}$ Work function $(\phi)=1.24 \mathrm{eV}$ Einstein's photoelectric equation- $\mathrm{eV}_{\mathrm{o}}=\frac{\mathrm{hc}}{\lambda}-\phi \quad\left(\because \mathrm{c}=3 \times 10^{8} \mathrm{~m} / \mathrm{s}\right)$ $\mathrm{eV}_{\mathrm{o}}=\frac{6.62 \times 10^{-34} \times 3 \times 10^{8}}{4.36 \times 10^{-7}}-1.24$ $\mathrm{~V}_{\mathrm{o}}=\frac{6.62 \times 10^{-34} \times 3 \times 10^{8}}{1.6 \times 10^{-19} \times 4.36 \times 10^{-7}}-1.24$ $\mathrm{~V}_{\mathrm{o}}=2.84-1.24$ $\mathrm{~V}_{\mathrm{o}}=1.60 \mathrm{eV}$
AP EAMCET (18.09.2020) Shift-I
Dual nature of radiation and Matter
142215
Radiation of two photons having energies twice and five times the work function of a metal are incident successively on the metal surface. Find out the ratio of maximum velocity of photo electrons emitted in the two cases.
1 $\mathrm{v}_{1} / \mathrm{v}_{2}=1 / 3$
2 $\mathrm{v}_{1} / \mathrm{v}_{2}=1 / 4$
3 $\mathrm{v}_{1} / \mathrm{v}_{2}=1$
4 $\mathrm{v}_{1} / \mathrm{v}_{2}=1 / 2$
Explanation:
D Given that, $\mathrm{E}_{1}=2 \phi$ $\mathrm{E}_{2}=5 \phi$ $\because \quad \mathrm{E}_{1}=\phi+\mathrm{K} . \mathrm{E}_{1}$ Putting the value of $E_{1}$, we get - $2 \phi-\phi=K . E_{1}$ $\phi=K . E_{1}$ $\text { And } \quad \mathrm{E}_{2}=\phi+\mathrm{KE}_{2}$ Putting the value of $\mathrm{E}_{1}$, we get - $5 \phi=\phi+\mathrm{KE}_{2}$ $\mathrm{KE}_{2}=4 \phi$ Dividing equation (i) by (ii), we get- $\therefore \quad \frac{\frac{1}{2} \mathrm{mv}_{1}^{2}}{\frac{1}{2} \mathrm{mv}_{2}^{2}}=\frac{1}{4}$ $\frac{v_{1}^{2}}{v_{2}^{2}}=\frac{1}{4}$ $\frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}=\frac{1}{2}$
SRMJEE-2019
Dual nature of radiation and Matter
142217
If frequency of a photon is $6 \times 10^{14} \mathrm{~Hz}$, then find its wavelength [ Take, speed of light, $\mathrm{c}=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$ ]
1 $500 \AA$
2 $500 \mathrm{~nm}$
3 $200 \AA$
4 $200 \mathrm{~nm}$
Explanation:
B Given that, $\mathrm{c}=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$ Frequency of photon $(v)=6 \times 10^{14} \mathrm{~Hz}, \lambda=?$ We know that, \(v=\frac{\mathrm{c}}{\lambda}\) Putting these value, we get- \(6 \times 10^{14}=\frac{3 \times 10^{8}}{\lambda}\) $\quad \lambda=0.5 \times 10^{-6}$ $\lambda=5 \times 10^{-7} \times \frac{100}{100}$ $\lambda=5 \times 100 \times 10^{-9} \mathrm{~m} \quad\left[\because 10^{-9} \mathrm{~m}=1 \mathrm{~nm}\right]$ $\lambda=500 \mathrm{~nm}$
