Light coming from a discharge tube filled with hydrogen falls on the c...
Energy of photon from H atom is gievn by the difference between the two energy level in which electron transit.for minimum wavelength, energy of the photon will be maximum.
maximum energy will be
according to the einstein equation,Maximum kinetic energy if the emitted electron = energy of the incident radiation − work function
KE = 13.6 − 4 = 9.6 eV
hence potenial of the node must be greater than or equal to 9.6 volt with negatice polarity.
Hnece option (d) is correct.
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Light coming from a discharge tube filled with hydrogen falls on the c...
To make the photo current zero, we need to ensure that no electrons are emitted from the cathode. This can be achieved by making the energy of the incident photons less than the work function of the cathode.
The energy of a photon is given by the equation E = hf, where E is the energy, h is Planck's constant, and f is the frequency of the light.
The work function of the cathode is 4eV. To convert this to joules, we multiply by the elementary charge, e, which is 1.6 x 10^-19 C. So, the work function in joules is 4e(1.6 x 10^-19) J.
Now, we can equate the energy of the photon to the work function:
hf = 4e(1.6 x 10^-19)
f = (4e(1.6 x 10^-19))/h
The anode voltage is the potential difference between the anode and the cathode. To make the photo current zero, we need to make this potential difference at least equal to the energy of the photons. This can be achieved by making the anode voltage equal to the energy of the photons, divided by the elementary charge:
Anode voltage = (hf)/e
Substituting the value of f from the previous equation, we have:
Anode voltage = [(4e(1.6 x 10^-19))/h]/e
Simplifying, we find:
Anode voltage = (4(1.6 x 10^-19))/h
Substituting the value of Planck's constant, h = 6.63 x 10^-34 J.s, we can calculate the anode voltage:
Anode voltage = (4(1.6 x 10^-19))/(6.63 x 10^-34)
Calculating this expression, we find:
Anode voltage ≈ 1.93 x 10^14 V
Therefore, the anode voltage that will likely make the photo current zero is approximately 1.93 x 10^14 Volts.