A light wave described by E = 60[sin(3 times 10^15t) + sin(12 times 10^15t)] (in SI units) falls on a metal surface of work function 2.8 eV. The maximum kinetic energy of ejected photoelectron is (approximately) ____ eV. (h = 6.6 times 10^-34text Jcdottexts. and e = 1.6 times 10^-19textC)

Solution & Explanation

### Related Formula K_textmax = hnu_textmax - phi_0 v = fracomega2pi ### Core Logic The light wave consists of two frequencies governed by omega_1 and omega_2. omega_1 = 3 times 10^15text rad/s omega_2 = 12 times 10^15text rad/s The maximum kinetic energy of ejected photoelectrons will be determined by the highest frequency photon, which corresponds to omega_2 = 12 times 10^15text rad/s. ### Step 1: Calculate Photon Energy Frequency nu_textmax = fracomega_22pi = frac12 times 10^152 times 3.14 approx 1.91 times 10^15text Hz. Energy of this photon: E_textphoton = hnu = (6.6 times 10^-34) times (1.91 times 10^15) = 1.26 times 10^-18text J Convert this energy to eV: E_textmax = frac1.26 times 10^-181.6 times 10^-19 approx 7.87text eV approx 7.9text eV ### Step 2: Calculate Maximum Kinetic Energy Using Einstein's photoelectric equation: K_textmax = E_textmax - phi_0 K_textmax = 7.9 - 2.8 = 5.1text eV ### Pattern Recognition When a wave has multiple frequency components (E = E_1sinomega_1 t + E_2sinomega_2 t), the K_textmax is always strictly determined by the highest frequency (highest energy) component. Ignore the lower frequency terms. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Dual Nature of Radiation and Matter

Reference Study Guides

Dual Nature Of Radiation And Matter

More Dual Nature of Radiation and Matter Previous-Year Questions — Page 4

Q jee_main_2025_29_jan_morning Photoelectric Effect
Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : Emission of electrons in photoelectric effect can be suppressed by applying a sufficiently negative electron potential to the photoemissive substance. Reason (R): A negative electric potential, which stops the emission of electrons from the surface of a photoemissive substance, varies linearly with frequency of incident radiation. In the light of the above statements, choose the most appropriate answer from the options given below:
  • A. text(A) is false but (R) is true.
  • B. text(A) is true but (R) is false.
  • C. textBoth (A) and (R) are true and (R) is the correct explanation of (A).
  • D. textBoth (A) and (R) are true but (R) is not the correct explanation of (A).

Solution

### Related Formula eV_0 = hnu - phi_0 ### Core Logic Assertion (A) is true because applying a negative stopping potential decelerates the emitted photoelectrons and drops the output current down to zero. Reason (R) is true because the stopping potential V_0 = left(fracheright)nu - fracphi_0e is linear with frequency nu. However, the linearity of V_0 vs frequency does not explain the physical mechanism behind why a negative potential stops electron emission. ### Chapter Mix Class 12 Physics: Dual Nature of Radiation and Matter
Q jee_main_2025_29_jan_morning de Broglie Wavelength
If lambda and K are de Broglie Wavelength and kinetic energy, respectively, of a particle with constant mass. The correct graphical representation for the particle will be :-
  • A.
  • B.
  • C.
  • D.

Solution

### Related Formula lambda = frachsqrt2mK lambda^2 = left(frach^22mright) frac1K ### Core Logic Rearranging the de Broglie equation displays a parabolic relationship when evaluating squared attributes or corresponding axes coordinates. Given standard lambda vs frac1sqrtK layout tracking, it exhibits an upward facing parabolic behavior matching option (2) layout. ### Chapter Mix Class 12 Physics: Dual Nature of Radiation and Matter
Q41 jee_main_2024_01_february_morning de Broglie Wavelength
The de Broglie wavelengths of a proton and an alpha particle are lambda and 2lambda respectively. The ratio of the velocities of proton and alpha particle will be:
  • A. 1 : 8
  • B. 1 : 2
  • C. 4 : 1
  • D. 8 : 1

Solution

### Related Formula de Broglie wavelength relationship to velocity: lambda = frachp = frachmv implies v = frachmlambda ### Core Logic Let mass of proton be m_p and mass of alpha particle be m_alpha = 4m_p. Given wavelengths: lambda_p = lambda, lambda_alpha = 2lambda. Set up ratios: fracv_pv_alpha = fracm_alpham_p times fraclambda_alphalambda_p ### Step 1: Substitute Ratios $fracv_pv_alpha = 4 times frac2lambdalambda = 4 times 2 = 8 Hence, the velocity ratio is 8:1. ### Pattern Recognition Remember the standard mass ratio: m_\alpha \approx 4 m_p$. Inversely proportional components mean velocity amplifies significantly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Dual Nature of Radiation and Matter
Q31 jee_main_2024_29_january_evening Photon Theory of Light
Two sources of light emit with a power of 200text W. The ratio of number of photons of visible light emitted by each source having wavelengths 300text nm and 500text nm respectively, will be:
  • A. 1:5
  • B. 1:3
  • C. 5:3
  • D. 3:5

Solution

### Related Formula The power P of a light source emitting n photons per second of wavelength lambda is given by: P = n cdot frachclambda where: * h is Planck's constant * c is the speed of light ### Core Logic Since both light sources emit with the same power (P = 200text W), we can relate the number of photons emitted per second for each wavelength: n_1 frachclambda_1 = n_2 frachclambda_2 Cancelling out the constant terms h and c, we get: fracn_1lambda_1 = fracn_2lambda_2 implies fracn_1n_2 = fraclambda_1lambda_2 Thus, the ratio of the number of photons emitted is directly proportional to their wavelengths. ### Step 1: Substitute the Values Given values: * lambda_1 = 300text nm * lambda_2 = 500text nm Substituting these values into the ratio equation: fracn_1n_2 = frac300500 = frac35 Thus, the ratio is 3:5. ### Pattern Recognition Shortcut: For equal power outputs, the photon emission rate n is directly proportional to the wavelength lambda. Therefore, ratio of photons n_1 : n_2 = lambda_1 : lambda_2 = 300 : 500 = 3:5 directly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Dual Nature of Radiation and Matter
Q44 jee_main_2024_27_jan_morning Photoelectric Effect
A convex lens of focal length 40text cm forms an image of an extended source of light on a photoelectric cell. A current I is produced. The lens is replaced by another convex lens having the same diameter but focal length 20text cm. The photoelectric current now is:
  • A. fracI2
  • B. 4I
  • C. 2I
  • D. I

Solution

### Core Logic Photoelectric current is directly proportional to the intensity of incident light, which depends on the amount of light energy intercepted. The amount of light energy collected by a lens depends strictly on its aperture diameter. Since both lenses share the exact same diameter, they gather the same total light flux from the source and direct it onto the active photoelectric cell matrix. Thus, the total incident power is invariant. ### Step 1: Conclusion Since the incident energy flux remains constant, the rate of emission of photoelectrons remains identical, meaning the current stays exactly I. ### Pattern Recognition Focal length alters spatial image sizing metrics, but raw aperture dimensions rule total power intercept profiles in optical flux systems. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Dual Nature of Radiation and Matter, Ray Optics

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