Wavelengths belonging to Balmer series lying in the range of 450 nm to 750 nm were used to eject photoelectrons from a metal surface whose work function is 2.0 eV. Find ( in eV ) the maximum kinetic energy of the emitted photoelectrons. `("Take hc = 1242 eV nm.")`

Answer

To find the maximum kinetic energy of the emitted photoelectrons when wavelengths from the Balmer series are used, we can follow these steps:

### Step 1: Identify the relevant wavelengths
The wavelengths of interest from the Balmer series that lie within the range of 450 nm to 750 nm are:
- For the transition from n=3 to n=2:
\[ \lambda = 654.55 \, \text{nm} \]
- For the transition from n=4 to n=2:
\[ \lambda = 486.84 \, \text{nm} \]

### Step 2: Determine the maximum energy photon
The maximum energy photon corresponds to the minimum wavelength. Therefore, we will use the wavelength of 486.84 nm for our calculations.

### Step 3: Calculate the energy of the photon
Using the formula for the energy of a photon:
\[ E = \frac{hc}{\lambda} \]
where \( h \) is Planck's constant and \( c \) is the speed of light. Given \( hc = 1242 \, \text{eV nm} \), we can substitute:
\[ E = \frac{1242 \, \text{eV nm}}{486.84 \, \text{nm}} \]

### Step 4: Perform the calculation
Calculating the energy:
\[ E = \frac{1242}{486.84} \approx 2.55 \, \text{eV} \]

### Step 5: Subtract the work function
The work function \( \phi \) of the metal is given as 2.0 eV. The maximum kinetic energy \( K_{\text{max}} \) of the emitted photoelectrons can be calculated using Einstein's photoelectric equation:
\[ K_{\text{max}} = E - \phi \]
Substituting the values:
\[ K_{\text{max}} = 2.55 \, \text{eV} - 2.0 \, \text{eV} = 0.55 \, \text{eV} \]

### Final Answer
The maximum kinetic energy of the emitted photoelectrons is:
\[ \boxed{0.55 \, \text{eV}} \]

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