The threshold frequency of a certain metal is `3.3xx10^(14)Hz`. If light of frequency `8.2xx10^(14)Hz` is incident on the metal, predict the cut off voltage for photoelectric emission. Given Planck's constant, `h=6.62xx10^(-34)Js`.

Explanation

To solve the problem of predicting the cut-off voltage for photoelectric emission, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values**:
- Threshold frequency, \( \nu_0 = 3.3 \times 10^{14} \, \text{Hz} \)
- Frequency of incident light, \( \nu = 8.2 \times 10^{14} \, \text{Hz} \)
- Planck's constant, \( h = 6.62 \times 10^{-34} \, \text{Js} \)
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