Dual Nature NEET 2012: Threshold Frequency Solution

What is the answer to the Dual Nature NEET 2012 hydrogen-radiation question?

The Dual Nature NEET 2012 hydrogen-radiation question gives option C: subtract the photoelectron’s maximum kinetic energy from the photon energy, then convert the remainder into threshold frequency. Converting the full photon energy produces option D, the incident frequency instead.

Light emitted when a hydrogen electron falls from its first excited level to its ground level strikes a photosensitive surface. The measured stopping potential is 3.57 V; the required quantity is the surface’s threshold frequency.

The supplied options are:

The question bank classifies this as hard, with an expected solving time of 90 seconds. These are question-bank labels, not student-performance measurements.

How do you find the energy of the emitted hydrogen photon?

The emitted photon carries 10.2 eV, which becomes the incident photon energy at the surface. Start with the hydrogen levels, not the stopping potential. The ground state is the lowest level; the first excited state is the next level above it, not the ground state:

The hydrogen energy-level formula gives:

The emitted photon energy is positive because the atom loses energy. Subtract the final atomic energy from the initial atomic energy:

The official solution writes this as:

$h\nu =10.2\text{eV}$ This is the incident photon energy, not the work function. The hydrogen transition determines what reaches the surface, not how much energy the surface requires to release an electron.

How does the stopping potential give the work function?

The work function is 6.63 eV: the supplied photon energy minus the maximum kinetic energy carried away by an electron. Keep both energies in electronvolts. Converting to joules adds an unnecessary step.

Einstein’s photoelectric equation separates the three energies:

The symbols for work function and stopping potential are:

The work function is the minimum energy needed for escape. The stopping potential determines the maximum kinetic energy of the emitted electrons: $V_s=3.57\text{V}$

Volts measure potential difference; electronvolts measure energy. Their numerical values match here because the particle has the electron’s charge magnitude.

Rearrange before substituting: $\varphi =h\nu -e{V}_s$

The full energy account is:

The stopping potential identifies the excess energy. It does not identify the minimum energy needed to free an electron.

How do you convert the work function into threshold frequency?

The threshold frequency comes from the 6.63 eV work function, giving option C. At threshold, a photon supplies just enough energy for escape, so the maximum kinetic energy is zero:

Use the supplied conversion between photon energy and frequency:

Following the official calculation:

Rounded to the option precision:

The positive stopping potential confirms that the incident photons supply energy beyond the work function. The threshold frequency must therefore be below the incident frequency, not equal to it.

Use that comparison as a physical check on the answer. Retain this recall chain rather than memorising the final option:

Why does confusing incident and threshold frequency produce option D?

Option D comes from converting the entire hydrogen photon energy into frequency. That arithmetic correctly calculates the incident radiation frequency, but answers the wrong physical question. The error is treating the whole photon energy as the work function:

The missing operation is subtracting the maximum kinetic energy before dividing by Planck’s constant: $K_{max}=3.57\text{eV}$

A solution that never uses the stated stopping potential has not accounted for the escaping electrons’ kinetic energy. Check for that subtraction before checking the frequency conversion.

Record this as a quantity-identification error, using NEET Mock Test Analysis: Errors, Repairs and Retests to plan the repair. Name the requested quantity before choosing the energy to convert.

Which three related questions check whether you understand the method?

These checks separate incident frequency, stopping potential and intensity. They are original practice questions based on the solved setup, not additional verified NEET PYQs. Use the same material throughout, so its work function stays unchanged.

What is the frequency of the incident hydrogen radiation?

Convert the full 10.2 eV into frequency because this question asks about the incoming photons. Unlike the threshold calculation, no kinetic energy is subtracted:

This is the incident frequency, not the material’s threshold frequency. The threshold uses only the energy needed for escape:

What is the new stopping potential if the incident frequency doubles?

The new stopping potential is 13.77 V, not twice the original value. Doubling frequency doubles photon energy to 20.4 eV, while the same material still requires 6.63 eV for escape:

The work function is subtracted once from each photon’s energy. It does not double with the incident frequency.

What happens if intensity doubles at unchanged frequency?

In the standard photoelectric-effect model, the stopping potential remains 3.57 V. Photon energy and work function are unchanged, so maximum kinetic energy is unchanged:

$V_s=3.57\text{V}$ Higher intensity supplies more photons per unit time, not more energy per photon. Before calculating any intensity variant, check whether the frequency changed.

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