Stopping Potential
Trending Questions
Define stopping potential.
- 2
- 4
- 6
- 10
Reason: Specific charge is the ratio of the charge to mass.
Both assertion and reason are true and the reason is the correct explanation of the assertion.
Both assertion and reason are true but reason is not the correct explanation of the assertion.
Assertion is true but reason is false.
Assertion and reason both are false.
- Assertion is false but reason is true.
- The stopping potential will be 0.2 V
- The stopping potential will be 0.6 V
- The saturation current will be 6.0 mA
- The saturation current will be 2.0 mA
- 1.8 V
- 1.3 V
- 0.5 V
- 2.3 V
Light of two different frequencies whose photons have energies 1 eV and 2.5 eV respectively, successively illuminate a metallic surface whose work function is 0.5 eV. Ratio of maximum speeds of emitted electrons will be
1:2
1:5
1:1
1:4
- 454 nm
- 113 nm
- 906 nm
- 226 nm
How is rms value of AC voltage related to peak value of AC voltage
[ Take mass of electron, me=9.1×10−31 kg ]
- 3×10−21 ms−1
- 2.7×10−21 ms−1
- 2.7×10−18 ms−1
- 3×10−18 ms−1
- 0.67×10−7 m/s
- 60 m/s
- 10.3×107 m/s
- 1.5×107 m/s
- 2V0
- V04
- V02
- V0
In an X-ray tube, electrons emitted from a filament (cathode) carrying current hit a target (anode) at a distance from the cathode. The target is kept at a potential higher than the cathode resulting in the emission of continuous and characteristic X-rays. If the filament current is decreased to , the potential difference is increased to , and the separation distance is reduced to , then
The cut-off wavelength will reduce to half, and the wavelengths of the characteristic X-rays will remain the same
The cut-off wavelength, as well as the wavelengths of the characteristic X-rays, will remain the same
The cut-off wavelength will reduce to half, and the intensities of all the X-rays will decrease
The cut-off wavelength will become two times larger, and the intensity of all the X-rays will decrease
An electron in a hydrogen atom undergoes a transition from an orbit with quantum number ni to another with quantum number nf. Vi and Vf are respectively the initial and final potential energies of the electron. If VfVi=6.25, then the smallest possible nf is
- >v(43)1/2
- <v(43)1/2
- =v(43)1/2
- =v(34)1/2
(i) In which case is the stopping potential more and why ?
(ii) Does the slope of the graph depend on the nature of the material used ? Explain.
- hce(1λ3+1λ2−1λ1]
- hce(1λ3−1λ2−1λ1]
- hce(1λ3+12λ2−1λ1]
- hce(1λ3+12λ2−32λ1]