चिकित्सा शिक्षा आयोग (CEE) बीएस्सी नर्सिङ र बीएनएस प्रवेश तयारी
चिकित्सा शिक्षा आयोग (CEE) – बीएस्सी नर्सिङ र बीएनएस प्रवेश
B.Sc. Nursing & BNS Entrance · Medical Education Commission – CEE
२,०३९ प्रश्न उपलब्ध। मोड छान्नुहोस्:
MBBS/BDS/Nursing प्रवेश: प्रश्न र उत्तर
सही लागेको विकल्पमा थिच्नुहोस्। सही भए हरियो, गलत भए रातो देखिन्छ; ३ सेकेन्डपछि सही उत्तर र व्याख्या देखिन्छ।
सही: ० / ०
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An n-type semiconductor is obtained by doping silicon with:
सही उत्तर: घ) Phosphorus
Phosphorus is pentavalent; its fifth valence electron becomes a free charge carrier. Trivalent impurities like boron, aluminium and gallium produce p-type semiconductors.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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1 atomic mass unit (u) is equivalent to an energy of approximately:
सही उत्तर: ग) 931.5 MeV
From E = mc², 1 u ≈ 931.5 MeV. 0.511 MeV is the rest energy of an electron.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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When a nucleus emits an alpha particle, its mass number and atomic number respectively:
सही उत्तर: ख) Decrease by 4 and by 2
An alpha particle is a helium nucleus (⁴₂He), so A drops by 4 and Z by 2; e.g. ²³⁸U → ²³⁴Th. The reverse pairing (2 and 4) is a common mix-up.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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In β⁻ (beta-minus) decay, the atomic number of the nucleus:
सही उत्तर: क) Increases by 1
A neutron converts into a proton, emitting an electron and an antineutrino, so Z increases by 1 while A is unchanged. In β⁺ decay Z decreases by 1.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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The ratio of the radius of the second Bohr orbit to that of the first orbit in a hydrogen atom is:
सही उत्तर: घ) 4 : 1
rₙ ∝ n², so r₂/r₁ = 2²/1² = 4. 2:1 would be true only if r were proportional to n.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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The Lyman series of the hydrogen spectrum lies in the:
सही उत्तर: ग) Ultraviolet region
Lyman lines are transitions ending on n = 1, which have large energies and fall in the ultraviolet. The Balmer series (ending on n = 2) lies in the visible region.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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The energy of the electron in the second orbit (n = 2) of a hydrogen atom is:
सही उत्तर: ख) −3.4 eV
Eₙ = −13.6/n² eV, so E₂ = −13.6/4 = −3.4 eV. −6.8 eV results from dividing by n instead of n²; −1.51 eV is the n = 3 level.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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The energy of a photon of wavelength 620 nm is approximately (take hc = 1240 eV nm):
सही उत्तर: क) 2.0 eV
E = hc/λ = 1240/620 = 2.0 eV. 3.1 eV corresponds to 400 nm (violet light).
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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In the photoelectric effect, if the intensity of incident light is increased keeping its frequency constant (above threshold), then:
सही उत्तर: घ) The number of photoelectrons emitted increases
Higher intensity means more photons per second, so more electrons are emitted, i.e. a larger photocurrent. The maximum kinetic energy and stopping potential depend only on frequency.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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The de Broglie wavelength of a particle of momentum p is:
सही उत्तर: ग) h/p
de Broglie proposed λ = h/p for matter waves. Hence a faster or heavier particle has a shorter wavelength.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
यी अभ्यास प्रश्न हुन्, आधिकारिक प्रश्न होइनन्। पाठ्यक्रम, विज्ञापन र नतिजाका लागि सम्बन्धित निकायको आधिकारिक वेबसाइट हेर्नुहोस्। उत्तरमा गल्ती भेटे प्रश्नमुनिको "टिप्पणी / गल्ती रिपोर्ट" मा लेख्नुहोस्।
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