Any radiation in the ultraviolet region of hydrogen spectrum is able to eject photoelectrons from a metal. What should be the maximum value of threshold frequency for the metal?

Options

A. `3.288times10^(15)Hz`
B. `2.466times10^(15)Hz`
C. `4.594times10^(14)Hz`
D. `8.220times10^(14)Hz`

Explanation

To solve the problem, we need to determine the maximum value of the threshold frequency for a metal that can eject photoelectrons when exposed to ultraviolet radiation from the hydrogen spectrum. Let's break this down step by step.

### Step 1: Understand the Concept of Threshold Frequency
The threshold frequency (\( \nu_0 \)) is the minimum frequency of light required to eject photoelectrons from a metal surface. If the frequency of the incident light is greater than this threshold frequency, photoelectrons will be emitted.

### Step 2: Identify the Relevant Series in the Hydrogen Spectrum
The ultraviolet region of the hydrogen spectrum primarily includes the Lyman series, which corresponds to electronic transitions from higher energy levels to the first energy level (n=1). The transition that gives the minimum frequency in the Lyman series is from \( n=2 \) to \( n=1 \).

### Step 3: Calculate the Energy Difference
The energy levels of the hydrogen atom are given by the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
For the transition from \( n=2 \) to \( n=1 \):
\[ E_2 = -\frac{13.6}{2^2} = -\frac{13.6}{4} = -3.4 \, \text{eV} \]
\[ E_1 = -\frac{13.6}{1^2} = -13.6 \, \text{eV} \]
The energy difference (\( \Delta E \)) for this transition is:
\[ \Delta E = E_1 - E_2 = (-13.6) - (-3.4) = -13.6 + 3.4 = -10.2 \, \text{eV} \]

### Step 4: Relate Energy to Frequency
The energy of a photon can also be expressed in terms of its frequency using the equation:
\[ E = h \nu \]
where \( h \) is Planck's constant (\( 6.63 \times 10^{-34} \, \text{Js} \)) and \( \nu \) is the frequency. Therefore, we can set:
\[ h \nu = 10.2 \, \text{eV} \]
To convert eV to joules, we use the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):
\[ 10.2 \, \text{eV} = 10.2 \times 1.6 \times 10^{-19} \, \text{J} \]

### Step 5: Solve for Frequency
Now we can solve for the frequency (\( \nu \)):
\[ \nu = \frac{10.2 \times 1.6 \times 10^{-19}}{h} \]
Substituting the value of \( h \):
\[ \nu = \frac{10.2 \times 1.6 \times 10^{-19}}{6.63 \times 10^{-34}} \]

### Step 6: Calculate the Frequency
Calculating the above expression:
\[ \nu \approx \frac{1.632 \times 10^{-18}}{6.63 \times 10^{-34}} \approx 2.466 \times 10^{15} \, \text{Hz} \]

### Conclusion
The maximum value of the threshold frequency for the metal is approximately:
\[ \nu \approx 2.466 \times 10^{15} \, \text{Hz} \]