Any radiation in the ultra violet region of Hydrogen spectrum is able to eject photo-electrons from a metal. Then the maximum value of threshold wave length is `lambda= (1)/(R) = 911 Å` then the frequency of the metal is, nearly

Options

A. `3.3xx10^(15)` HZ
B. `2.5xx10^(15)` HZ
C. `4.6xx10^(14)` HZ
D. `8.2xx10^(14)` HZ

Answer

To solve the problem, we need to find the frequency of the metal based on the given threshold wavelength. Here are the steps to arrive at the solution:

### Step 1: Understand the relationship between wavelength and frequency
The frequency (ν) of electromagnetic radiation is related to its wavelength (λ) by the equation:
\[ ν = \frac{c}{λ} \]
where:
- \( c \) is the speed of light, approximately \( 3 \times 10^8 \) m/s.

### Step 2: Convert the threshold wavelength from Angstroms to meters
The given threshold wavelength is:
\[ λ = 911 \, \text{Å} \]
To convert Angstroms to meters, we use the conversion factor \( 1 \, \text{Å} = 10^{-10} \, \text{m} \):
\[ λ = 911 \times 10^{-10} \, \text{m} \]

### Step 3: Substitute the values into the frequency equation
Now, we can substitute the values into the frequency equation:
\[ ν = \frac{c}{λ} = \frac{3 \times 10^8 \, \text{m/s}}{911 \times 10^{-10} \, \text{m}} \]

### Step 4: Perform the calculation
Calculating the above expression:
\[ ν = \frac{3 \times 10^8}{911 \times 10^{-10}} \]
\[ ν = \frac{3 \times 10^8}{9.11 \times 10^{-8}} \]
\[ ν \approx 3.29 \times 10^{15} \, \text{Hz} \]

### Step 5: Round to significant figures
Rounding the result gives:
\[ ν \approx 3.3 \times 10^{15} \, \text{Hz} \]

### Conclusion
Thus, the frequency of the metal is nearly \( 3.3 \times 10^{15} \, \text{Hz} \), which corresponds to the first option.

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