| ⇦ | 
| ⇨ |
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: - Work of 3.0×10⁻⁴ joule is required to be done in increasing the size of a soap film
- The half-life period of a radio-active element X is same as the mean life time
- When we close one slit in the young’s double slit experiment, then
- The nuclear fusion reaction between deutrium and tritium takes place
- Two closed organ pipes when sounded simultaneously give 4 beats/second.
Topics: Dual Nature of Matter and Radiation
(150)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Work of 3.0×10⁻⁴ joule is required to be done in increasing the size of a soap film
- The half-life period of a radio-active element X is same as the mean life time
- When we close one slit in the young’s double slit experiment, then
- The nuclear fusion reaction between deutrium and tritium takes place
- Two closed organ pipes when sounded simultaneously give 4 beats/second.
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