Dual Nature of matter and Radiation Part 2

Concept: The photoelectric effect relates the energy of incident light (frequency) to the work function and the maximum kinetic energy of the emitted photoelectrons.
Key Rule/Formula: The energy of a photon is E=hν=hc∕λ. The photoelectric equation is Kmax=hν−φ, or Kmax=hc∕λ−φ.
Solution/Analysis:
1. According to the photoelectric effect, the maximum kinetic energy (K) of the emitted photoelectrons is given by K=hν−φ, where φ is the work function and h is Planck's constant.
2. For light of wavelength λ1, the energy is E1=hc∕λ1. Thus, K1=hc∕λ1−φ.
3. For light of wavelength λ2, the energy is E2=hc∕λ2. Thus, K2=hc∕λ2−φ.
4. We are given the relationship between the wavelengths: λ1=2λ2. This implies ‌

1

λ1

=‌

1

2λ2

.
5. Substitute this relationship into the expression for K1:
K1=hc∕λ1−φ
K1=hc∕(2λ2)−φ
6. Now, we use the expression for K2:
K2=hc∕λ2−φ
We can rewrite hc∕λ2 as 2(hc∕(2λ2)). Substituting this into the equation for K2:
K2=2(hc∕(2λ2))−φ
7. Substitute the expression from step 5 into the equation for K2:
K2=2(K1+φ)−φ (This substitution is complex; let's use a simpler approach based on the structure of the provided solution.)
Alternative Step-by-Step using the provided logic:
1. From step 2 and 3, we have:
K1=hc∕λ1−φ‌‌(i)
K2=hc∕λ2−φ‌‌(ii)
2. Using the given condition λ1=2λ2, we substitute λ1 in equation (i):
K1=hc∕(2λ2)−φ‌‌(iii)
3. Rearrange equation (iii) to find hc∕(2λ2):
K1+φ=hc∕(2λ2)
4. Now, substitute this expression into equation (ii):
K2=(hc∕λ2)−φ
Since hc∕λ2=2(hc∕(2λ2)), we get:
K2=2(K1+φ)−φ
K2=2K1+2φ−φ
K2=2K1+φ
5. Rearranging the terms to find φ:
φ=K2−2K1