JEE Mains · Chemistry · STD 11 - 2. structure of atom
The work functions of two metals ( \(M_A\) and \(M_B\) ) are in the \(1: 2\) ratio. When these metals are exposed to photons of energy 6 eV , the kinetic energy of liberated electrons of \(M_A: M_B\) is in the ratio of \(2.642: 1\). The work functions (in eV ) of \(M _{ A }\) and \(M _{ B }\) are respectively.
- A 3.1, 6.2
- B 2.3, 4.6
- C 1.4, 2.8
- D 1.5, 3.0
Answer & Solution
Correct Answer
(B) 2.3, 4.6
Step-by-step Solution
Detailed explanation
\( KE_{max} = E - \phi \) \( (KE_{max})_{1} = 6 - \phi_{1} \)\(\quad\) ... (1) \( (KE_{max})_{2} = 6 - \phi_{2} \) \(\quad\)... (2) By eq. (1) divide eq. (2) \(\frac{\left( KE _{\max }\right)_1}{\left( KE _{\max }\right)_2}=\frac{2.642}{1}=\frac{6-\phi_1}{6-\phi_2}\)…
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