Tiêu đề 11. DUAL NATURE OF RADIATION AND MATTER.pdf ✅

11. DUAL NATURE OF RADIATION AND MATTER NEET PREPARATION (MEDIUM PHYSICS PAPER) Date: March 12, 2025 Dura on: 1:00:00 Total Marks: 180 INSTRUCTIONS INSTRUCTIONS PHYSICS 1. () : Explana on de-Broglie wavelength λ = h p Also, E = p 2 2m ⇒ p = √2mE ∴ λ = h √2mE (E = kine c energy). So correct op on is (3) 2. () : Explana on hv1 − hv0 = 1 2mv 2 1 hv 2 − hv 0 = 1 2mv 2 2 ∴ h v1 − v2 = 1 2m v 2 1 − v 2 2 ∴ v1 = f1 and v2 = f2 ∴ v 2 1 − v 2 2 = 2h m f1 − f2 3. () : Explana on Given : Frequency, v = 6 × 1014 Hz; Power, P = 2 × 10 − 3 W As, P = nhv t So, number of photons emi ed by the source per sec, n t = P hv = 2 × 10 − 3 6.63 × 10 − 34 × 6 × 1014 = 0.050 × 1017 = 5 × 1015 4. () : Explana on Photons have zero rest mass and always travel at the speed of light. No material body can ex‐ ceed this speed. So, if two photons move in opposite direc ons, their rela ve velocity to each other remains the speed of light. So cor‐ rect op on is (2). 5. () : Explana on From Einstein's equa on for photoelectric emission, maximum kine c energy of photo electrons is given by KE max = E − ω0 = E − E 3 = 2E 3 ∴ KE lies from 0 to 2E 3 . Correct op on is (3). ( ) ( ) [ ] ( )
6. () : Explana on According to ques on, Mass of electron = m Wavelength of electron = λ Poten al difference in case of electron = V From de - Brogile rela on, λ = h p ⇒ h = h √2mKE = h √2mqV ⇒ λ ∝ 1 √qVm For electron, λ ∝ 1 √eVm (1) For proton, λ1 = 1 √e9VM (2) Where, e is the charge on proton, poten al differnce = 9 V Mass of proton = m From eq.(1) and (2), we get λ λ1 = 9VMe eVm ⇒ λ1 = λ 3 m M 7. () : Explana on λ = h mv where, m = m0 √1 − v 2 / c 2 As v = c, then ⇒ m = m0 1 − c 2 c 2 = m0 √1 − 1 = m0 0 ⇒ m = ∞, Hence, λ = h ∞ = 0 8. () : Explana on Satura on current depends upon the intensity of light which is inversely propor onal to the square of the distance of the lamp from the photosensi ve surface. Thus i ∝ 1 d 2 9. () : Explana on Stopping poten al equals to maximum kine c energy. Since stopping poten al is varying lin‐ early with the frequency, therefore max. KE for both the metals also vary linearly with fre‐ quency. As x and y intercepts are different then threshold frequency, work func on and stopping poten als are different for both the metals. 10. () : Explana on For photo electric emission wavelength (λ) of radia on must be less than threshold wave‐ length (λ0 ). ω0 = 2.4 eV λ0 = 12400 ω0 = 12400 2.4 λ0 = 5166A 0 = 516.6nm = maximum ∴ For λ = 700 nm photo electric effect does not take place. Correct op on is (3). 11. () : Explana on Given, K = kine c energy, m = mass Then, K = 1 2mv 2 (1) But we know that, λ = h mv (2) Now, from Eq. (1), v 2 = 2K m ⇒ v = √2K/m Pu ng this in Eq. (2), we get = h √2mK √ √ √
12. () : Explana on T1 = 300K, λ1 T2 = 600K, λ 2 = ? de-Broglie wavelength is given by λ = h P = h √2mKE = h 2m × 3 2 KT Here KE = Average KE of each molecule = 3 2 KT. λ ∝ 1 √T ⇒ λ2 λ1 = T1 T2 = 300 600 = 1 √2 or λ2 = λ1 √2 13. () : Explana on Conceptual Ques on 14. () : Explana on The kine c energy of emi ed photoelectrons varies from zero to a maximum value. Work func on depends on nature of metal. Op on (3) is correct. 15. () : Explana on The specific charge of a posi ve ion corre‐ sponding to one gas is fixed but it is different for different gases. The specific charge depends on charge and mass of posi ve ions present in posi ve rays So correct op on is (2). 16. () : Explana on As, mv = h λ (de Broglie equa on) or = h mλ = 6.6 × 10 − 34 9.1 × 10 − 31 × 5200 × 10 − 10 = 1.4 × 103 ms − 1 17. () : Explana on power (P) = n hc λ n = pλ hc = 3.3 × 10 − 3 × 6 × 10 − 7 6.6 × 10 − 34 × 3 × 108 = 1016 18. () : Explana on Op on (2) is correct. 19. () : Explana on Efficient power (P) is given by P = N t × hc λ N = total number of photons N t = P× λ hc = 50 × 0.6 × 10 − 6 6.6 × 10 − 34 × 3 × 10 8 = 1.5 × 1020 [ ∵ 25% of 200 W = 50 W] 20. () : Explana on By Einstein's photoelectric equa on, KE max = E − φ = 4 ⋅ 13 − 3.13 ⇒ K. E max = 1 eV 21. () : Explana on Electron microscopes achieve superior resolv‐ ing power due to their use of electrons with shorter wavelengths compared to op cal mi‐ croscopes using visible light. Op cal wave‐ length are from 4500 ∘ A to 7000 ∘ A . But elec‐ tron is a ma er wave (when moving) with wavelength λ = h me ve . Given, ve = 5 × 105 m/s ⇒ λe = 6.6 × 10 − 34 9 × 10 − 31 × 5 × 105 ⇒ λe = 0.147 × 10 − 8 m = 1.47 nm = 14.7 ∘ A This is very much less than the lower limit 450 nm of op cal visible region. Then, by Rayleigh's criterion, Resolving power = d 1.22λ ∝ 1 λ ⇒ Electron microscope has very high resolv‐ ing power compared to op cal device. So cor‐ rect op on is (1). 22. () : Explana on Stopping poten al is the poten al where the current is zero. ∴ − 8v √ √ √ ( ) ( )