To solve this problem, let's use Einstein's photoelectric equation:
[tex]\[ K.E = h (f - f_0) \][/tex]
where:
- [tex]\( K.E \)[/tex] is the kinetic energy of the emitted photoelectrons,
- [tex]\( h \)[/tex] is Planck's constant,
- [tex]\( f \)[/tex] is the frequency of the incident light,
- [tex]\( f_0 \)[/tex] is the threshold frequency for the metal.
Given:
- Frequency [tex]\( f_1 = 0.4 \times 10^{13} \text{ Hz} \)[/tex]
- Frequency [tex]\( f_2 = 1.0 \times 10^{13} \text{ Hz} \)[/tex]
- The kinetic energy of photoelectrons emitted with frequency [tex]\( f_1 \)[/tex] is twice that of those emitted with frequency [tex]\( f_2 \)[/tex], i.e., [tex]\( K.E_1 = 2 \times K.E_2 \)[/tex]
From the photoelectric equation, we have:
[tex]\[ K.E_1 = h (f_1 - f_0) \][/tex]
[tex]\[ K.E_2 = h (f_2 - f_0) \][/tex]
Given [tex]\( K.E_1 = 2 \times K.E_2 \)[/tex]:
[tex]\[ h (f_1 - f_0) = 2 \times h (f_2 - f_0) \][/tex]
We can simplify this by dividing through by Planck's constant [tex]\( h \)[/tex]:
[tex]\[ f_1 - f_0 = 2 (f_2 - f_0) \][/tex]
Expanding the right-hand side:
[tex]\[ f_1 - f_0 = 2f_2 - 2f_0 \][/tex]
Rearranging terms to isolate [tex]\( f_0 \)[/tex]:
[tex]\[ f_1 + f_0 = 2f_2 \][/tex]
[tex]\[ f_0 = 2f_2 - f_1 \][/tex]
Substituting the given values of [tex]\( f_1 \)[/tex] and [tex]\( f_2 \)[/tex]:
[tex]\[ f_0 = 2 \times (1.0 \times 10^{13}) - (0.4 \times 10^{13}) \][/tex]
[tex]\[ f_0 = 2.0 \times 10^{13} - 0.4 \times 10^{13} \][/tex]
[tex]\[ f_0 = 1.6 \times 10^{13} \, \text{Hz} \][/tex]
Therefore, the threshold frequency for the metal is:
[tex]\[ \boxed{1.6 \times 10^{13} \text{ Hz}} \][/tex]
So the correct answer is (C) [tex]\(1.6 \times 10^{13} \text{ Hz}\)[/tex].
