| ⇦ | 
| ⇨ |
A light whose frequency is equal to 6.0 x 10¹⁴ Hz is incident on a metal whose work function is 2 eV. (h= 6.63 x 10⁻³⁴ Js, 1 ev = 1.6 x 10⁻¹⁹ J). The maximum energy of the electron emitted will be
Options
(a) 2.49 eV
(b) 4.49 eV
(c) 0.49 eV
(d) 5.49 eV
Correct Answer:
0.49 eV
Explanation:
No explanation available. Be the first to write the explanation for this question by commenting below.
Related Questions: - One-fourth length of a spring of force constant K is cut away. The force constant
- Rutherford’s α-scattering experiment concludes that
- Identify the paramagnetic substance
- Which of the two have same dimensions?
- Two parallel long wires carry currents i₁ and i₂ with i₁ > i₂ .
Topics: Dual Nature of Matter and Radiation
(150)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- One-fourth length of a spring of force constant K is cut away. The force constant
- Rutherford’s α-scattering experiment concludes that
- Identify the paramagnetic substance
- Which of the two have same dimensions?
- Two parallel long wires carry currents i₁ and i₂ with i₁ > i₂ .
Topics: Dual Nature of Matter and Radiation (150)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends
Leave a Reply